Potential

01 拉普拉斯方程和他在笛卡尔坐标系下的解

拉普拉斯方程其人

开局先写一个假设什么东西都不动的麦克斯韦方程组

$\nabla E = \frac{\rho}{\epsilon_0}$

$\nabla B = 0$

$\nabla \times E = 0$

$\nabla \times B = \mu_0 J$

$E = -\nabla V$ ，把这个带入第一个式子就可以得到著名的泊松equation

$\nabla^2V = -\frac{\rho}{\epsilon_0}$

而如果在一个范围内没有任何电荷出现的话那就可以得到另一个叫做Laplace's equation的东西

$\nabla^2 V = 0$

还挺优雅的不是嘛

来解方程吧！用一个具体的例子！

$X(x)Y(y)Z(z)$ 的。我们可以举一个具体的例子来熟悉这种方法

20231013_143402496_iOS

$\pm$ $V = V_0$ ；然后y=0和a的那两个面是接地的。这也就是说Z(z)这一项没有什么用不妨把他设成1.所以我们现在有

$V = X(x)Y(y)$

把他带入拉普拉斯方程中

$0=\frac{\partial^2V}{\partial x^2}+\frac{\partial^2V}{\partial y^2}$

就可以得到

$Y\frac{\partial^2 X}{\partial x^2}+X\frac{\partial^2 Y}{\partial y^2}=0$

两边同时除V

$\frac{1}{X}\frac{\partial^2 X}{\partial x^2}+\frac{1}{Y}\frac{\partial^2 Y}{\partial y^2}=0$

因为第一项只和x有关，第二项只和另一个有关，所以只有这俩都是常数的情况，而且偏微分可以写成全微分

先尝试首项是负数的情况

$\frac{1}{X}\frac{d^2 X}{d x^2}=-\frac{1}{Y}\frac{d^2 Y}{d y^2}=-k^2$

$X(x) = A_1\cos (kx)+A_2\sin (kx)$ $Y(y)=B_1e^{ky}+B_2e^{-ky}$ $Y(0)=Y(a)=0$ $B_1+B_2=B_1e^{ka}+B_2e^{-ka}=0$ $B_1(e^{ka}-e^{-ka})=0$ $2B_1\sinh(ka)=0$ $B_1=B_2=0$ 这就是说所有地方的V都是0，这显然不是个解，也就是说我们要尝试一下首项是正数的情况

$\frac{1}{X}\frac{d^2 X}{d x^2}=-\frac{1}{Y}\frac{d^2 Y}{d y^2}=k^2$

$Y(y) = A_1\cos (ky)+A_2\sin (ky)$ $X(x)=B_1e^{kx}+B_2e^{-kx}$ （没错我就换了一下xy，甚至连顺序都懒得换）我们有

$Y(0)=0$

$A_1=0$ $Y(a)=A_2 \sin(ka)=0$ $A_2$ $\sin(ka)=0$ $ka=n\pi$ $A_2$ 是随机常数。所以最终我们的解为

$V(x,y)=\overset{n}{\underset{n=0}{\sum}}C_n\cosh(\frac{n\pi x}{a})\sin(\frac{n\pi y}{a})$

$C_n$ 怎么搞？答案是用X(x)的边界条件

$V(b,y)=\overset{n}{\underset{i=0}{\sum}}C_n\cosh(\frac{n\pi x}{a})\sin(\frac{n\pi y}{a})=V_0$

$\sin(\frac{m\pi y}{a})$ ，再对y从0到a积分，可以得到

$V(b,y)=\overset{n}{\underset{i=0}{\sum}}C_n\cosh(\frac{n\pi x}{a})\overset{a}{\underset{0}{\int}}\sin(\frac{n\pi y}{a})\sin (\frac{m\pi y}{a})dy=V_0\overset{a}{\underset{0}{\int}}sin (\frac{m\pi y}{a})dy$

其中

$\sin \frac{n\pi y}{a}\sin \frac{m\pi y}{a}=\frac{1}{2}\left[\cos\frac{(n-m)\pi y}{a}-\cos\frac{(n+m)\pi y}{a}\right]$

$\frac{a}{2}$ $C_m=0$ $C_m=\frac{4V_0}{m\pi \cosh\left(\frac{m\pi b}{a}\right)}$

最终我们的解就是一个特别恐怖的式子

$V(x,y)=\frac{4V_0}{\pi}\overset{n}{\underset{i=0}{\sum}}\frac{\cosh \frac{(2i+1)\pi x}{a}}{\cosh \frac{(2i+1)\pi b}{a}}{a}\sin(\frac{i\pi y}{a})$

结束！

02 拉普拉斯方程在球坐标下的解

优雅 $\nabla^2V=0$ 并且算了算他在笛卡尔坐标系下的一个例子，这次来试一试球坐标

球坐标下的拉普拉斯方程

通过一些矢量微积分，我们可以知道

$\nabla^2V = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial V}{\partial r}\right) + \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial V}{\partial \theta}\right) + \frac{1}{r^2 \sin^2\theta}\frac{\partial^2 V}{\partial \phi^2} = 0$

嗯。。。以我现在的知识还是看不出来这有什么优雅的

解方程！

一杯茶，一包烟，一道积分算一天！（虽然我不抽烟）

$\nabla^2=R(r)\Theta(\theta)\Phi(\phi)$

$\frac{r^2}{R\Theta\Phi}$ ，可以得到

$\frac{1}{R}\frac{\partial}{\partial r}\left(r^2 \frac{\partial R}{\partial r}\right) + \frac{1}{\Theta \sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial \Theta}{\partial \theta}\right) + \frac{1}{\Phi \sin^2\theta}\frac{\partial^2 \Phi}{\partial \phi^2} = 0$

$\sin^2\theta$ $\theta$ $\phi$ $\sin^2 \theta$ 是一个常数：

$\frac{1}{\Phi}\frac{d^2\Phi}{d^2\phi}=K$

$e^{\pm\sqrt{K}\phi}$ $\phi$ $\phi=\phi_0+2n\pi$ $\phi$ $-m^2$ 。这样子的话我们就可以得到

$\Phi(\phi)=Ce^{im\phi}$

$\phi$ 可以是正数也可以是负数但必须要是整数。求个导的话就是

$\frac{d^2\Phi}{d\phi^2}=-m^2\Phi$

整个的式子就变成了这样子

再找找有没有什么只和一个变量相关的东西？

（没错！）

（是第一项！）

（嘿嘿！）

那我们就可以把第一项设成常数了！

$\frac{1}{R}\frac{\partial}{\partial r}\left(r^2 \frac{\partial R}{\partial r}\right) =l(l+1)$

$\frac{1}{\Theta \sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta \frac{\partial \Theta}{\partial \theta}\right) - \frac{m^2}{\sin^2\theta} = l(l+1)$

至于为什么把常数设计得这么怪，你之后就知道啦

$\cos\theta$ 设为x，那么下面的式子就变成了

$\frac{d}{dx}\left[(1-x^2)\frac{d\Theta(x)}{dx}\right]+\left[l(l+1)-\frac{m^2}{1-x^2}\right]\Theta(\theta)=0$

苍天啊大地啊我都干什么啊！这玩意怎么越写越长啊

别怕，这个方程在之后的数学物理方法中就可以学到其实叫做Legendre equation。他的解叫做associated Legendre polynomials，可以写成这种形式

$\Theta(x)=P^m_l(x)$

$l$ 是正整数，不然的话这个方程就又不是周期性的了，而且m也只能是从-l到+l

现在知道为啥把常数设成l(l+1)了吧

$\Theta$ $\Phi$ spherical harmonic $Y^m_l$ 表示

$Y_l^m=CP_l^m(\cos\theta)e^{im\phi}$

然后就是这个式子了

$\frac{1}{R}\frac{\partial}{\partial r}\left(r^2 \frac{\partial R}{\partial r}\right) =l(l+1)$

求一下导

$l(l+1)=2\frac{r}{R}\frac{dR}{dr}+\frac{r^2}{R}\frac{d^2R}{dr^2}$

$R=Cr^x$ ，也就是说

$l(l+1)=\frac{2}{Cr^{x-1}}Cxr^{x-1}+\frac{1}{Cr^{x-2}}Cx(x-1)r^{x-2}$

$l(l+1)=2x+x(x-1)$

$l(l+1)=x(x+1)$

$l$ $-l-1$

所以说R就出来了

$R(r)=\alpha r^l+\frac{\beta}{r^{l+1}}$

把三个东西乘起来就可以得到

$V(r,\theta,\phi)=\overset{\infty}{\underset{l=0}{\sum}}\overset{l}{\underset{m=-l}{\sum}}\left(\alpha r^l+\frac{\beta}{r^{l+1}}\right)Y^m_l(\theta,\phi)$

Legendre多项式的一个特点

P有一个很有用的特点，就是他们是正交的，也就是说我们有

$\int^1_{-1}P_n(x)P_m(x)dx=\frac{2}{2n+1}\delta_{m,n}$

$\alpha$ $\beta$

一个对称的情况

例子我喜欢例子现在就让我算例子！！

$\phi$ $\cos\theta$ 的Legendre多项式，也就是说整个的V可以写成

$V(r,\theta)=\overset{\infty}{\underset{l=0}{\sum}}\left(\alpha r^l+\frac{\beta}{r^{l+1}}\right)P_l(\cos \theta)$

我们可以举一个例子：在一个三维空间里面，存在着一个空心导体球在原点，他的半径是a。同时还存在一个沿着z轴向上的匀强电场E，求电势

首先啊，这是一个球，所以我们肯定会用球坐标来解这个方程

其次我们看一下示意图

$\phi$ 没有什么关系！那我们就可以用那个简化版的拉普拉斯方程的解

$V(r,\theta)=\overset{\infty}{\underset{l=0}{\sum}}\left(\alpha r^l+\frac{\beta}{r^{l+1}}\right)P_l(\cos \theta)$

$-E\cos\theta$ 。我们看一下Legendre多项式的前几个方程：

$P_0(x)=1，P_1(x)=x，P_2(x)=\frac{1}{2}(3x^2-1)，P_3(x)=\frac{1}{2}(5x^3-3x)\cdots$

$\alpha$ $\beta$ 才不等于0

所以说就可以得到

$V(r,\theta)=\left(-E r+\frac{\beta_1}{r^{2}}\right)\cos \theta$

还是一样的end behavior

就快要算出来了！

$\beta_1$ ，我们还有另外一个边界条件没有用，也就是r=a的情况:

$V(a,\theta)=(-Ea+\frac{\beta_1}{a^2})\cos\theta=0$

$\beta_1=a^3E$

综上所述，我们能算出来

$V(r,\theta)=-Er\cos \theta+\frac{a^3E\cos\theta}{r^2}$

可以看出来其中第一项对应的是电场对应的电势，而后面哪一项就是磁场对应的电势了

完事收工！

Formal solution to Poisson's equation

Poisson's equation:

${\nabla}^2V=-\frac{\rho}{\epsilon_0}$ $V(\overrightarrow{r})=\frac{1}{4\pi\epsilon_0}\int\frac{\rho({\overrightarrow{r'}})}{|\overrightarrow{r'}-\overrightarrow{r}|}d\tau'$

$\tau'$ is the volume

$\overrightarrow{B} = \overrightarrow{\nabla} \times \overrightarrow{A}$ , and A is called magnetic vector potential

This is valid because the divergence of B is 0

$\overrightarrow{\nabla}\times(\overrightarrow{\nabla}\times A)=\mu_0\overrightarrow{J}$

$\overrightarrow{\nabla}(\overrightarrow{\nabla}\cdot A)-\overrightarrow{\nabla}^2\overrightarrow{A}$

$\overrightarrow{\nabla}\cdot A=0$

$\Rightarrow{\nabla^2}\overrightarrow{A} = -\mu_0\overrightarrow{J}$

it looks like three independent Poisson equations

$\Rightarrow \overrightarrow{A}(\overrightarrow{r})=\frac{\mu_0}{4\pi}\int\frac{\overrightarrow{J}(\overrightarrow{r'})}{|\overrightarrow{r'}-\overrightarrow{r}|}d\tau'$

Example: If we have a one dimensional current i.e. wire, this current will produce

$\overrightarrow{A}(\overrightarrow{r}) = \frac{\mu_0I}{4\pi}\int\frac{\overrightarrow{dl'}}{|\overrightarrow{r'}-\overrightarrow{r}|}$ $\overrightarrow{B}(\overrightarrow{r})=\overrightarrow{\nabla}\times\overrightarrow{A}=\frac{\mu_0}{4\pi}\int\frac{\overrightarrow{dl'}\times(\overrightarrow{r'}-\overrightarrow{r})}{|\overrightarrow{r'}-\overrightarrow{r}|^3}$ Which is Biot-Savart law!

Multipole expansion

Laplace with azimuthal symmetry:

$V(r,\theta)=\sum_{l=0}^\infty[A_lr^l+\frac{B_l}{r^{l+1}}]P_l(cos\theta)$

$P_l$ is Legendre polynomial

$\overrightarrow{r'}$

$V(\overrightarrow{r})=\frac{1}{4\pi\epsilon_0}\frac{q}{|\overrightarrow{r}-\overrightarrow{r'}|}$

(unit charge)

This has azimuthal symmetry, so we can use the above equation

$\gamma$ $cos\theta=1$ $P_l(cos\theta)=1$ , then

$V(r,\theta)=\frac{1}{4\pi\epsilon_0}\frac{q}{|\overrightarrow{r'}-\overrightarrow{r}|}=\sum^\infty_{l=0}A_lr^l+\frac{B_l}{r^{l+1}}$

$P_l$ (1)=1

The expansion would be

$\frac{1}{|\overrightarrow{r'}-\overrightarrow{r}|}=\left\{ \begin{matrix} \frac{1}{r}\sum^\infty_{l=0}(\frac{r'}{r})^l & r>r'\\ \\ \frac{1}{r'}\sum^\infty_{l=0}(\frac{r}{r'})^l & r<r' \end{matrix} \right.$

$r'<r$

The general solutiuon would be

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}\sum^\infty_{l=0}(\frac{r'}{r})^lP_l(cos\theta)$

$\overrightarrow{r}$ outside a charge distribution is

$V(\overrightarrow{r})=\frac{1}{4\pi\epsilon_0}\int_{l=0}^\infty(r')^lP_l(\cos(\gamma))\times\rho(\overrightarrow r')d\tau'$

The leading term is the monopole term (l=0)

$V(\overrightarrow{r})=\frac{1}{4\pi\epsilon_0}\int\rho(r')d\tau'=\frac{1}{4\pi\epsilon_0}\frac{Q}{r}$

$\frac{1}{r}$

Next is the dipole term (l=1)

$V_1(\overrightarrow{r})=\frac{1}{4\pi\epsilon_0}\int(r')\cos(\gamma)\rho(r')d\tau'$ $=\frac{\overrightarrow p\cdot\overrightarrow r}{4\pi\epsilon_0r^3}$

$\frac{1}{r^2}$

$\overrightarrow p$ $=\int\overrightarrow r'\rho(r')d\tau'$

$\frac{\overrightarrow r\cdot\overrightarrow r'}{rr'}=\cos(\gamma)$

Then the quadrupole term (l=2)

$V_2(\overrightarrow{r})=\frac{1}{4\pi\epsilon_0}\int(r')^2\frac{3\cos^2(\gamma)-1}{2}\rho(r')d\tau'$

Now we can have a nice way of looking at field outside the charges using multipole expansion.

2. Electric polarization

Matter is not free space, it contains atoms
The first two pictures are from Griffith Book

2.1 Bound charges

Electric field can induce a dipole moment in a neutral atom

$\mathbf p = \alpha \mathbf E$ $\alpha$ is the polarizability of the atom. A list of polarizability can be seen from the handout.

$10^6$ $10^{-15}$ m, which is much less the size of an atom.

$K_B T$

However, an E field can line them up

$\mathbf P$ $\mathbf P = n\mathbf p$

If they are not aligned, then the polarization is zero.

Question: what is the electric potential produced by a box of dipoles?

\begin{aligned} V (R) & = \frac{1}{4 π ϵ_{0}} \int \frac{P (r) \cdot R}{| R |^{3}} d^{3} τ^{'} \\ = \frac{1}{4 π ϵ_{0}} \int P (r) \cdot \nabla \frac{1}{R} d τ^{'} \\ \nabla \frac{1}{R} & = - \frac{R}{R^{3}} \\ \nabla \frac{1}{| r - r^{'} |} & = - \frac{r - r^{'}}{| r - r^{'} |^{3}} \\ \nabla^{'} \frac{1}{| r - r^{'} |} & = \frac{r - r ‘}{| r - r^{'} |^{3}} \end{aligned}

$\mathbf\nabla\cdot (f\mathbf F)=f\mathbf\nabla\cdot\mathbf F+\mathbf F\cdot\mathbf\nabla f$

\Rightarrow V (r) = \frac{1}{4 π ϵ_{0}} \int \nabla^{'} \frac{P (r^{'})}{| R |} d τ^{'} - \frac{1}{4 π ϵ_{0}} \int \frac{1}{R} \nabla \cdot P (r^{'}) d τ^{'}

= \frac{1}{4 π ϵ_{0}} \int_{s^{'}} \frac{σ_{b} d S^{'}}{R} + \frac{1}{4 π ϵ_{0}} \int_{τ^{'}} \frac{ρ_{b} d τ^{'}}{R}

$\sigma_b = \mathbf P\cdot\mathbf n$ $\rho_b = -\nabla\cdot\mathbf P$ $\mathbf P$ is defined as the dipole moment per unit volume).

The derivation can be seen from Griffiths' book page 176.

Remember to drop primes!

$\nabla\cdot\mathbf E = \frac{\rho}{\epsilon_0}$ $\rho = \rho_{free} + \rho_{bound}$ $\rho_{bound} = -\nabla\cdot\mathbf P$ $\nabla\cdot\mathbf E = \frac{\rho_{free}}{\epsilon_0} - \frac{\nabla\cdot\mathbf P}{\epsilon_0}$

We define electric displacement

D = ϵ_{0} E + P

\Rightarrow \nabla \cdot D = ρ_{f r e e}

That is much easier, since it only depends on free charges.

The integral form is

\oint D \cdot d a = Q_{f r e e}

Linear dielectrics

$\mathbf P = \epsilon_0\chi_e\mathbf E$ $\chi_e$ is the electric susceptibility, which is dimensionless.

\Rightarrow D = ϵ_{0} E + P = ϵ_{0} (1 + χ_{e}) E

$(1+\chi_e)$ $\epsilon_r$ and it is dimensionless.

Example:

Field from a point charge in a linear dielectric:

\int_{s} D \cdot d s = Q_{f r e e}

\Rightarrow D = \frac{Q_{f}}{4 π r^{2}} \hat{r}

\Rightarrow E = \frac{Q_{f}}{4 π ϵ_{0} r^{2}} \hat{r}

$\overrightarrow{P}=n\overrightarrow{p}$ $\rho_b=-\nabla\cdot\overrightarrow{P}$ $\sigma_b=\overrightarrow{P}\cdot\hat{n}$ $\overrightarrow{D}=\epsilon_0\overrightarrow{E}+\overrightarrow{P}=\epsilon_0\epsilon_r\overrightarrow{E}$ $\nabla\cdot\overrightarrow{D}=\rho_f$ $\overrightarrow{P}=\epsilon_0\chi_e\overrightarrow{E}$ $\chi_e$ $\epsilon_r-1$

Question Has D solved everything? -- No. The div equation is easier, but...

$\overrightarrow{\nabla}\cdot\overrightarrow{E}=\frac{\rho_{tot}}{\epsilon_0}$ $\overrightarrow{\nabla}\cdot\overrightarrow{D}=\rho_f$ $\overrightarrow{D}$ $\epsilon_0\overrightarrow{E}+\overrightarrow{P}=\epsilon_0\epsilon_r\overrightarrow{E}$

\vec{\nabla} \times \vec{E} = 0

\vec{\nabla} \times \vec{D} = ϵ \vec{\nabla} \times \vec{E} + \vec{\nabla} \times \vec{P} \neq 0

Boundrary conditions

$E_{//}$ is continuous

\vec{\nabla} \times \vec{E} = 0

\oint \vec{E} \cdot d \vec{l} = 0

$D_{//}$ $\nabla\times\overrightarrow P\ne0$ at interface)

$D_\perp$ is continuous

\nabla \cdot \vec{D} = ρ_{f}

\oint \vec{D} \cdot d \vec{a} = Q_{f}

$E_\perp$ may not be continuous if we have bound charges at the interface

Example

\Rightarrow E_{0} / / x

E_{1} = E_{3} = E_{0}

$D_\perp$ is continuous

\Rightarrow D_{1} = D_{3} = ϵ_{0} E_{0} = D_{2}

$D_2=\epsilon_0\epsilon_rE_2$

\Rightarrow E_{2} = \frac{E_{0}}{ϵ_{r}}

So there is a smaller E field in the dielectric

\begin{matrix} P_{2} = ϵ_{0} (ϵ_{r} - 1) E_{2} \\ = \frac{ϵ_{0}}{ϵ_{r}} (ϵ_{r} - 1) E_{0} \end{matrix}

$(P_1=P_3=0)$

$P_2$ $\nabla\times\overrightarrow P=0\Rightarrow \rho_b=0$

$z = o, \sigma_b = \mathbf P\cdot\mathbf n = \frac{\epsilon_0}{\epsilon_r}(\epsilon_r-1)E_0$ $z = d, \sigma_b = \mathbf P\cdot\mathbf n = \frac{\epsilon_0}{\epsilon_r}(\epsilon_r-1)E_0$

Example:

$\mathbf{E_0}$ . Find V.

We got to use Laplace's equation

General solution to Laplace's equation in spherical coordinates (with azimuthal symmetry) is

$V(r,\theta) = \sum^{\infty}_{l = 0}\left(A_lr^l+\frac{B_l}r^{l+1}\right)P_l(\cos\theta)$

$P_l$ is the Legendre polynomial.

Solve separatively for r > a and r < a

$r\rightarrow\infty$ $V\rightarrow -E_0r\cos \theta$ $r\rightarrow 0$ $V$ $\Rightarrow B_l = 0$ for r<a

$V = V_{in}\sum^{\infty}_{l = 0}A_lr^lP_l(\cos\theta)$

$V = V_{out}=-E_0r\cos\theta+\sum^{\infty}_{l = 0}\frac{B_l}{r^{l+1}}P_l(\cos\theta)$

$E_{//}, D_{\perp}, V$ are continuous

$\sum_{l=0}^{\infty}A_lr^lP_l(\cos \theta) = -E_0 a \cos\theta+\sum_{l=0}^\infty\frac{B_l}{a^{l+1}}P_l(\cos\theta)$

$\Rightarrow A_l = \left\{ \begin{matrix}\frac{B_l}{a^{l+1}}&&l\ne 1\\ -E_0+\frac{B_1}{a^2} && l=1 \end{matrix}\right.$

$E_0 = -\frac{1}{r}\frac{\partial V}{\partial \theta}$ is continuous

$\rightarrow$ generates same expressions

That is not terribly useful, so we will try another one

$D_{\perp} = -\epsilon_0\epsilon_r\frac{\partial V}{\partial r}$ is continuous

$\Rightarrow -\epsilon_0\epsilon_r\frac{\partial V}{\partial r}|_{r=a} = -\epsilon_0\frac{\partial V_{out}}{\partial r}|_{r=a}$

$\epsilon_r\sum_{l=0}^\infty lA_la^{l-1}P_l(\cos\theta) = -E_0\cos\theta+\sum_{l=0}^\infty -(l+1)\frac{B_l}{a^{l+2}}P_l(\cos\theta)$

$A_l = \left\{\begin{matrix}-\frac{1}{\epsilon_r}\frac{l+1}{l}\frac{B_l}{a^{2l+1}} && l\ne 1 \\ \frac{1}{\epsilon_r}(-E_0-\frac{2B_l}{a^3})&&l=1\end{matrix}\right.$

$A_l=B_l=0$ $l\ne 1$

On the other hand, we have

$A_1 = \underbrace{-E_0+\frac{B_1}{a^3}}_①=\underbrace{\frac{1}{\epsilon_r}\left(-E_0-\frac{2B_0}{a^3}\right)}_②$

$\epsilon_r(①-②) = 0$

Hence

$(1-\epsilon_r)E_0+(\epsilon_r+2)\frac{B_1}{a^3}=0$

$\Rightarrow B_1 = a^3E_0\left(\frac{\epsilon_r-1}{\epsilon_r+2}\right)$

$\rightarrow A_1 = -\frac{3E_0}{\epsilon_r+2}$

Now, put everything together

$V = \left\{\begin{matrix} \underbrace{\frac{3}{\epsilon_r+2}E_0r\cos\theta}_{\text{uniform field}} && r<a\\ \underbrace{-E_0r\cos\theta}_{\text{uniform field}}+\underbrace{\left(\frac{E_r-1}{E_r+2}\frac{E_0a^3\cos\theta}{r^2}\right)}_{\text{dipole field}}&&r>a\end{matrix}\right.$

Interestingly, we all have uniform fields applied inside and outside. And if in vacuum, we will have uniform field outside (dipole field vanishes)

P inside sphere:

$P = \epsilon_0(\epsilon_r-1)E_{in}$

$E_in$ $\frac{3E_0}{\epsilon_r+2}$

Dipole moment of sphere

$p= \frac{4}{3}\pi a^3P=\frac{4\pi a^3\epsilon_0(\epsilon_r-1)E_0}{\epsilon_r+2}$

Polarizability

$\begin{align*}\alpha &= \frac{P}{E_0}\\ &=4\pi a^3\epsilon_0\frac{\epsilon_r-1}{\epsilon_r+2}\end{align*}$

Bound charge on the surface

$\begin{align*}\sigma_b&=\mathbf P\cdot\hat{\mathbf n}\\&=P\cos\theta\\* &=\frac{3\epsilon_0(\epsilon_r-1)}{\epsilon_r+2}E_0\cos\theta\end{align*}$

3 Magnetization

3.1 Current Loop

From the handout, we have the following vector identity:

$\oint \mathbf{\hat{r}\cdot r'}d\mathbf{l'} = \left(\int d\mathbf{S'}\right )\times \mathbf{\hat{r}}$

Start from Poisson's equation:

$\nabla^2 A = \frac{-\mu_0}{4\pi} \int \left(\frac{\mathbf{J}(\mathbf{r'})}{|\mathbf{r}-\mathbf{r'}|}\right)d\mathbf{\tau'}$

Current loop:

$A(r) = \frac{\mu_0I}{4\pi} \oint \frac{1}{|\mathbf{r}-\mathbf{r'}|}d\mathbf{l'}$

recall the expansion

$\frac{1}{|\mathbf{r}-\mathbf{r'}|} = \frac{1}{r}\sum_{l=0}^{\infty}\left(\frac{r'}{r}\right)^l P_l(\cos\theta)$

$\Rightarrow A(r) = \frac{\mu_0I}{4\pi r}\sum_{l=0}^{\infty}\left(\frac{r'}{r}\right)^l \oint P_l(\cos\theta)d\mathbf{l'}$

l = 0 term is zero

$A_0(r) = \frac{\mu_0I}{4\pi r}\oint d\mathbf{l'} = 0$

n = 1 term is the next most important term

$A_1(r) = \frac{\mu_0I}{4\pi r}\left(\frac{1}{r}\right)\oint r' \cos\theta d\mathbf{l'}$

$r' \cos\theta = \mathbf r'\cdot\mathbf{\hat r}$ hence

$A_1(r) = \frac{\mu_0I}{4\pi r^2}\oint \mathbf r'\cdot\mathbf{\hat r} d\mathbf{l'}=\frac{\mu_0I}{4\pi r^2}\mathbf m\times\mathbf{\hat{r}}$
$\begin{align*}\mathbf m &= I\int_sd\mathbf{S'} \\ &= I\mathbf{S'}\end{align*}$

Is the magnetic dipole moment of the current loop.

Lets stop the expansion before we get too complex stuff.

And that looks like a dipole right?

$\mathbf{m}\parallel z$ in spherical coordinates:

$\mathbf A_1(\mathbf r) = \frac{\mu_0}{4\pi r^2} m\sin(\theta)\hat{\mathbf \phi}$

$\mathbf B(\mathbf r) = \nabla\times\mathbf A = \frac{\mu_0}{4\pi r^3} m\sin(\theta)\hat{\mathbf r}-\frac{\mu_0}{4\pi r^2} m\cos(\theta)\hat{\mathbf \theta}$

$\frac{P}{4\pi\epsilon_0}\rightarrow\frac{\mu_0m}{4\pi}$

Field patterns look the same at large r, but at small r, the magnetic field is not singular.

3.2 Magnetic Properties

$\mathbf B$ magnetization $\mathbf{M} = n\mathbf m$ $\mathbf m$ is the magnewtic moment of one atom

There are three main effects:

$\rightarrow \mathbf{M} ∝ -\mathbf B$ and that effect is very weak

All materials exihibits it!

$\rightarrow \mathbf{M} ∝ +\mathbf B$ and it's stronger

$\mathbf{M} ∝ \mathbf B$ $CuSO_4\cdot5H_2O$

Ferromagnetism

Very strong QM effect E.g. Fe, Co, Ni It is a non-linear effect of M(B)

Recap to lecture 7:

$A(\mathbf r) = \frac{\mu_0}{4\pi r^2} \mathbf m\times\hat{\mathbf \phi}$ $\mathbf m = I\mathbf{S'}$ $\mathbf M = n\mathbf m$ $\mathbf m$ is the magnewtic moment of one atom

3.3 Field due to a box of magnetic dipoles

When doing this, we start with some vector identities:

start with Divergence theorem:

$\int_V \nabla\cdot\mathbf{u}d\tau = \oint_S \mathbf{u}\cdot d\mathbf{\tau}$ $\mathbf{u} = \mathbf v\times\mathbf c$ $\mathbf c$ is a constant vector

$\int_s \mathbf v\times\mathbf c\cdot d\mathbf S = \mathbf c\cdot\int_s \mathbf v\times d\mathbf S = 0$ $\mathbf v\times d\mathbf S$ $\mathbf c$

$\int_\tau\mathbf\nabla\cdot(\mathbf v\times\mathbf c)d\tau = \int_\tau\mathbf c\cdot(\nabla\times\mathbf v)d\tau - \int_\tau\mathbf v\cdot(\underbrace{\nabla\times\mathbf c}_0)d\tau$

$\mathbf c$ $\Rightarrow \int_S\mathbf v \times d \mathbf S = -\int_\tau\nabla\times\mathbf vd\tau$

$\mathbf r'$

$\mathbf A(\mathbf r) = \frac{\mu_0}{4\pi}\int \frac{\mathbf{M(r)}\times(\mathbf{r-r'})}{|\mathbf r-\mathbf r'|^3}d\tau'$

which equals

$\frac{mu_0}{4\pi}\int \mathbf{M(r)}\times\nabla\frac{1}{|r-r'|}d\tau'$

We have

$\nabla'\times\left(\mathbf{\frac{M}{r-r'}}\right)=\left(\frac{1}{|r-r'|}\mathbf\nabla'\times\mathbf M - \mathbf M \times \mathbf \nabla'\frac{1}{|r-r'|}\right)$

Hence

$A(\mathbf r) = \frac{\mu_0}{4\pi}\int\frac{\nabla'\times\mathbf{M(r)}}{|\mathbf{r-r'}|}-\frac{\nabla'\times\mathbf{M(r)}}{|\mathbf{r-r'}|}d\tau'$

The last term can be transformed into a surface integral:

$\mathbf{A(r)} = \frac{\mu_0}{4\pi}\int\frac{\nabla'\times\mathbf{M(r)}}{|\mathbf{r-r'}|}d\tau' + \frac{\mu_0}{4\pi}\oint\frac{\mathbf{M(r)}\times d\mathbf{S'}}{|\mathbf{r-r'}|}$

$\frac{\mu_0}{4\pi}\int_\tau\frac{\mathbf{J(r)}}{|r-r'|}d\tau+\frac{\mu_0}{4\pi}\int_s\frac{\mathbf K_b(\mathbf r')}{|r-r'|}dS'$

$\mathbf J = \nabla\times\mathbf M$ $\mathbf K_b = \mathbf M\times\mathbf{\hat n}$ is the surface bound current density.

Remember to drop the prime!

Having done this, we can fix up Amperes law:

$\nabla\times\mathbf B = \mu_0(\mathbf J_f+\mathbf J_b)$

$\mathbf J_f$ $\mathbf J_b$ is the bound current density, it can be written as the curl of M

$\mathbf J_f = \nabla\times(\mathbf{\frac{B}{\mu_0}}-\mathbf M)$

Stuff in the bracket is defined as the magnetic field H and hence the free current density has a simpler form

$\mathbf J_f = \nabla\times\mathbf H$

Its integral form is

$\oint\mathbf H\cdot d\mathbf l = I_{f}$

$\nabla\cdot B = 0$ $\nabla\cdot H = -\nabla\cdot M$

The equtions to remember would be

$\mathbf D = \epsilon_0\mathbf E + \mathbf P$ $B = \mu_0(\mathbf{H+M})$

Linear materials

$\mathbf M = \chi_m\mathbf H$ $\chi_m$ is the magnetic susceptibility

$\mathbf B = \mu_0(\mathbf H+\mathbf M) = \mu_0\underbrace{(1+\chi_m)}_{\mu_r}\mathbf H = \mu_0\mu_r\mathbf H$

$\mu_r = 1+\chi_m$ is the relative permeability

If willing to be strict, H would be called magnetic field strength and B would be called magnetic flux density

3.4 Boundary conditions

one is not changed:

$\nabla\cdot\mathbf B = 0 \Rightarrow B_1^{\perp} = B_2^{\perp}$

Making this cylinder flater and flater, we can see that the B field is continuous

$H_1^{\perp} = H_2^{\perp}$ $\nabla \cdot H = -\nabla \cdot M$

$\nabla\times \mathbf H = \mathbf J_f \Rightarrow H_1^{\parallel} - H_2^{\parallel}$ (Assuming there is no surface free currents)

$B^\parallel$ $\nabla\times\mathbf B = \mu_0(\mathbf J_f+\mathbf J_b)$

$\Rightarrow E^\parallel, D^\perp,H^\parallel, B^\perp \text{are continuous}$

Assuming there is no surface free charges/currents

3.5 Magnetic scalar potential

$\mathbf J_f = 0$ everywhere, then

$\nabla\times\mathbf H = 0 \Rightarrow \mathbf H = -\nabla\phi_m$

$\phi_m$ is the magnetic scalar potential

If, we are dealing wit linear material,

$\mathbf B = -\mu_0\mu_r\mathbf \nabla\phi_m$

$\nabla\cdot H = 0$

$\phi_m$

3.6 Ferromagnetism

$\lim_{\delta v\rightarrow 0}\frac{\sum_{i\in \delta v} m_i}{\delta v}$ $\ne$ 0, even in B = 0 The reason for ferromagnetism is the (what will be learnt next term), the exchange interaction Only in certain materials: Fe, Co, Ni, Gd

There is an energy cost due to the stray field of the magnetic dipoles.

From the picture having stray field, it can be seen that on the tip and bottom of the field, there is a divergence of magnitization, leading to divergence in H. Meaning that field spreads out whenever you have magnetic moment going into the surface and not flowing out.

The energy cost would be

$\int\frac{B^2}{2\mu_0}d\tau$ $\Rightarrow$ It is energetic flavorable to form domains

This will cost low energy since there is no stray field and hence no divergence in H. (Average over the surface, M is zero)

When applying a B field, the correlated domain is expanded and finally they are all aligned.

So, the magnetization process involves moving domain walls.

$\Rightarrow$ $\Rightarrow$ hysterisis loop States that is interesting

Saturation (right up)
Remanence (middle up)
Coercivity (left middle) Better picture is in handout

The field which is positive but M is zero is called coercive field.

We can modle this as

$M = (\mu_r-1)H$ $\mu_r = \mu_r(H)$

$\mu_r(H)$ is a multivalued function (because it depends on histroy)

Hard materials

$H_c$ $M_r$ are large

difficult to move domain walls
So it is hard to magnetize and demagnetize
so it is used for permanent magnets

Soft materials

$H_c$ $M_r$ are small

easy to move domain walls
so it is used for transformers, motors, etc.

Example: Magnetization of a ring Iron ring, radius r, and current I with N terms used to magnetize the ring

$\oint\mathbf H\cdot d\mathbf l = NI$ Since H only depends on the free current, it is favorable for us to use it.

Continue the calculation

$\begin{align*} H\cdot 2\pi r&= NI\\ H &= \frac{NI}{2\pi r}\end{align*}$

$\mu_0\mu_rH = \mu_0\mu_r\frac{NI}{2\pi r}$

Now lets put a gap in the ring

The cutout, x, would much less than r

Lets use the same trick, ampere's law

$\oint\mathbf H\cdot d\mathbf l = NI$

$H_{core}(2\pi r-x) + H_{gap}x = NI$

$B_{\perp}$ $\Rightarrow$ $B_{core} = B_{gap}$ $\frac{B_{core}}{\mu_0\mu_r}(2\pi r-x) + \frac{B_{gap}}{\mu_0}(x) = NI$

$B_{gap} = \frac{\mu_0\mu_rNI}{(2\pi r - x)+\mu_r x}$

$\mu_r >> \frac{2\pi r}{x}$ $B_{gap}$ $\frac{\mu_0NI}{x}$

$\mu_rNI$ to be very large.

4 Electromagnetic waves in material

Let's let everything move!

4.1 Displacement current

Conservation of charge

\nabla \cdot J_{s} = - \frac{\partial ρ_{f}}{\partial t}

However, this is incompatible with

\nabla \times H = J_{f}

Lets take the divergence for both sides, we get

\nabla \cdot (\nabla \times H) = 0 = \nabla \cdot J + \frac{\partial}{\partial t} \underset{ρ_{f}}{\underset{⏟}{\nabla \cdot D}}

$\Rightarrow$ $\mathbf{J_f} = \mathbf{\nabla\times H}$

\Rightarrow \nabla \times H = J_{f} + \underset{displacement current}{\underset{⏟}{\frac{\partial D}{\partial t}}}

$\mathbf{\nabla\times \left(\frac{B}{\mu_0}-M\right)}=\frac{1}{\mu_0}\mathbf{\nabla\times B-\underbrace{\nabla\times M}_{J_b}}$

$\mathbf{J_b}$ is the bound current density.

$=\mathbf{J_s}+\frac{\partial \mathbf{D}}{\partial t}=\mathbf{J_s}+\frac{\partial \mathbf{P}}{\partial t}+\epsilon_0\frac{\partial \mathbf{E}}{\partial t}$

$\Rightarrow$ $\mathbf{\nabla\times B}=\mu_0\left(\mathbf{J_f+J_b+J_p}\right)+\mu_0\epsilon_0\frac{\partial\mathbf E}{\partial t}$

$J_p$ $\frac{\partial\mathbf P}{\partial t}$

$\mathbf{\nabla\cdot J_p} = -\frac{\partial \rho_p}{\partial t}$ from conservation of charge.

$\mathbf P$

4.2 Maxwell's equations in insulating linear dielectrics

$\mathbf J_f = 0$ $\mathbf J_b = 0$

Hence, we could get Maxwell's equation

\begin{matrix} \nabla \cdot D = 0 & \nabla \times E = - \frac{\partial B}{\partial t} \\ \nabla \cdot B = 0 & \nabla \times H = \frac{\partial D}{\partial t} \end{matrix}

$\mathbf D = \epsilon_0\epsilon_r\mathbf E+\mathbf P$ $\mathbf B = \mu_0\mu_r\mathbf H$

which gives

\begin{matrix} \nabla \cdot E = 0 & \nabla \times E = - \frac{\partial B}{\partial t} \\ \nabla \cdot B = 0 & \nabla \times B = μ_{0} μ_{r} ϵ_{0} ϵ_{r} \frac{\partial E}{\partial t} \end{matrix}

Consider

\nabla \times (\nabla \times E) = \nabla (\nabla \cdot E) - \nabla^{2} E = - \frac{\partial}{\partial t} \nabla \times B

\Rightarrow \nabla^{2} E = \underset{\frac{1}{v^{2}}}{\underset{⏟}{μ_{0} μ_{r} ϵ_{0} ϵ_{r}}} \frac{\partial^{2} E}{\partial t^{2}}

which is wave equation.

$\Rightarrow v = \frac{c}{n}$ $\frac{1}{\sqrt{\epsilon_0\mu_0}}$ $\sqrt{\epsilon_r\mu_r}$ where n is also called refractive index

Plane waves solutions

$\mathbf{\frac{\partial \Psi}{\partial x}} = \mathbf{\frac{\partial \Psi}{\partial y}} = 0$

$\mathbf{\nabla\cdot E}=0\Rightarrow \frac{\partial E_z}{\partial z} = 0$ $\mathbf{\nabla \cdot B}=0\Rightarrow\frac{\partial B_z}{\partial z} = 0$

$\mathbf{\nabla\times E} = -\frac{\partial \mathbf B}{\partial t}\Rightarrow \frac{\partial B_z}{\partial t} = 0$

$\mathbf{\nabla\times B} = \mu_0\mu_r\epsilon_0\epsilon_r\frac{\partial \mathbf E}{\partial t}\Rightarrow \frac{\partial E_z}{\partial t} = 0$

$E_z$ $B_z$ are constant in z and t, they are not part of wave motion

now analyze the x,y components of curl:

- \frac{\partial E_{y}}{\partial z} = - \frac{\partial B_{x}}{\partial t}, - \frac{\partial B_{y}}{\partial z} = \frac{1}{v^{2}} \frac{\partial E_{x}}{\partial t}

$\Rightarrow E_x, B_y$ are solutions

Lets then take

\frac{\partial^{2} E_{x}}{\partial t^{2}} = \frac{1}{v^{2}} \frac{\partial^{2} E_{x}}{\partial t^{2}}

\Rightarrow E_{x} (z, t) = E_{x 0} e^{i (\pm k z - ω t)} \hat{x}

Then, we could have

\Rightarrow B (z, t) = B_{0} e^{i (\pm k z - ω t)} \hat{y}

$\pm \mathbf z$ direction

\begin{aligned} \Rightarrow \mp k E_{0} & = - ω B_{0} \\ and \pm k B_{0} & = \frac{ω}{v^{2}} E_{0} \\ \Rightarrow \frac{E_{0}}{B_{0}} & = \pm \frac{ω}{k} = \pm v \end{aligned}

Define Impedance Z as

Z = | \frac{E_{0}}{H_{0}} | = \sqrt{\frac{μ_{0} μ_{r}}{ϵ_{0} ϵ_{r}}}

$H_0 = \frac{B_0}{\mu_0\mu_r}$

$v = -\int \mathbf E \cdot d\mathbf l$ $I = \oint \mathbf H \cdot d\mathbf l$

So dimension would work

$\epsilon_r = \mu_r = 1$ $Z = \sqrt{\frac{\mu_0}{\epsilon_0}} = 377\Omega$

$\mathbf{\nabla\times E = -\dot B}$ $\propto e^{i(\mathbf{k\cdot r}-\omega t)}$ , we could get

\begin{aligned} i k \times E & = - (- i ω) B \\ = i ω μ_{0} μ_{r} H \\ \Rightarrow Z & = | \frac{E}{H} | = \sqrt{\frac{μ_{0} μ_{r} ω}{k}} \end{aligned}

$v = \frac{c}{n} = \frac{\omega}{k} = \frac{1}{\sqrt{\epsilon_0\epsilon_r\mu_0\mu_r}}$

which is this wave

4.3 conductors

$\begin{align*}\mathbf{\nabla\cdot J_f} &= -\frac{\partial \rho_f}{\partial t}\\ \mathbf{\nabla\cdot J_p} &= -\frac{\partial \rho_p}{\partial t}\\ \mathbf{\nabla\cdot J_b} &= 0\end{align*}$ $\mathbf{\nabla\cdot(\nabla\times M)} = 0$

For conductors, we have

$\rho_f = 0$ $\mathbf J_f = \sigma \mathbf E$ $\sigma$ $\mathbf D = \epsilon_0\epsilon_r\mathbf E$ $\mathbf B = \mu_0\mu_r\mathbf H$ from linearity

Then we could get Maxwell's equation in conductors

$\begin{matrix} \mathbf{\nabla\cdot E} = 0 && \mathbf{\nabla\times E} = -\frac{\partial\mathbf B}{\partial t}\\ \mathbf{\nabla\cdot B} = 0 &&\textcolor{brown}{\mathbf{\nabla\times B} = \underbrace{\mu_0\mu_r\sigma\mathbf{E}}_{\text{conduction } \mathbf J} + \underbrace{\mu_0\mu_r\sigma_0\sigma_r\frac{\partial \mathbf E}{\partial t}}_{\text{dispalcement }\mathbf J}} \end{matrix}$

$\tau$ , and it is easy to prove (said Blundell)

$\mathbf\nabla\cdot\mathbf J = \frac{\partial \rho}{\partial t}$

$\mathbf\nabla\cdot\mathbf J$ $\sigma\cdot\mathbf{\nabla\cdot E}$ from Ohm's law

$\mathbf{\nabla\cdot E}$ $\frac{\rho}{\epsilon_0\epsilon_r}$ from Gauss's law

$\Rightarrow \rho(f) = \rho(0)e^{-\frac{t}{\tau}}$

$\tau = \frac{\epsilon_0\epsilon_r}{\sigma}$

$\tau$ $\rho(f)$ is very small.

$\omega$ $\frac{1}{\omega}$ $\tau$ :

Condition	Conductor Type	Charge Response	Conductivity
$\omega\tau\ll 1$	Good conductor	Charges respond very quickly	$\sigma\gg \epsilon_0\epsilon_r\omega$ Conduction current dominates
$\omega\tau\gg 1$	Bad conductor	Charges respond very slowly	$\sigma\ll \epsilon_0\epsilon_r\omega$ Displacement current dominates

Take real life examples

	$\sigma(\Omega m)$	$\epsilon_r$	$\frac{\sigma}{\epsilon_0\epsilon_r} (S^{-1})$
metal	$10^7$	1	$10^{19}$
Silicon	$4\cdot 10^{-4}$	11.7	$10^{7}$
Glass	$10^{-10}$	5	$10$

$5\cdot 10^{14}$ Hz

Let's now do some electromagnetism

$\begin{align*}\mathbf{\nabla\times(\nabla\times E)} &= \underbrace{\mathbf{\nabla(\nabla\cdot E)}}_{0} - \nabla^2\mathbf E\\ & = -\frac{\partial}{\partial t}\mathbf{\nabla\times B}\\ \nabla^2 \mathbf E &= \mu_0\mu_r\sigma\frac{\partial \mathbf E}{\partial t} + \mu_0\mu_r\sigma_0\sigma_r\frac{\partial^2 \mathbf E}{\partial t^2} \end{align*}$

$\mathbf{E,B}\perp$ to each others

$\Rightarrow \mathbf E = \mathbf E_0\hat{\mathbf{x}}e^{i(\mathbf{\widetilde k\cdot z}-\omega t)}$

$\mathbf{\widetilde k = k}+i{\mathbf\kappa}$

$\Rightarrow \widetilde k^2 = i\underbrace{\mu_0\mu_r\sigma\omega}_\mu + \underbrace{\mu_0\mu_r}_\mu\underbrace{\epsilon_0\epsilon_r}_{\epsilon}\omega^2$

$\begin{align*} (k+i\kappa)^2 &= k^2-\kappa^2+2ik\kappa\\ k^2-\kappa^2 &= \mu_0\epsilon_0\omega^2\\ 2k\kappa &= \mu\sigma\omega\Rightarrow k = \frac{\mu\sigma\omega}{2\kappa}\\ 0 &= \left(\frac{\mu\sigma\omega}{2}\right)^2\frac{1}{\kappa^2} - \kappa^2 - \mu\epsilon_0\omega^2\\ 0 &= (\kappa^2)^2 + \mu\epsilon\omega^2(\kappa^2)-\left(\frac{\mu\sigma\omega}{2}\right)^2\\ \kappa^2 &= -\frac{\mu\epsilon\omega^2}{2}\pm\sqrt{\left(\frac{\mu\epsilon\omega^2}{2}\right)^2+\left(\frac{\mu\sigma\omega}{2}\right)^2}\\ \kappa^2 &= \frac{\mu\epsilon\omega^2}{2}\left[\pm\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^2}-1\right] \end{align*}$

$\Rightarrow \kappa = \sqrt{\frac{\mu\epsilon}{2}}\omega\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^2}-1}$

Sub into original equation

$k = \frac{\mu\sigma\omega}{2\kappa} = \sqrt{\frac{\mu\epsilon}{2}}\omega\sqrt{\sqrt{1+\left(\frac{\sigma}{\epsilon\omega}\right)^2}+1}$ $\Rightarrow \mathbf E = \mathbf E_0\hat{\mathbf{x}}\underbrace{e^{-\kappa z}}_{e^{-\frac{z}{\delta}}}e^{i(\mathbf{k\cdot r}-\omega t)}$

$\delta = \frac{1}{\kappa}$ is the skin depth

Reminder $\sigma \gg \epsilon\omega$

$k = \kappa = \sqrt{\frac{\mu\epsilon}{2}}\omega\sqrt{\frac{\sigma}{\epsilon\omega}} = \sqrt{\frac{\mu\omega\sigma}{2}}$

We could therefore have

$\nabla^2\mathbf E = \mu\sigma\frac{\partial \mathbf E}{\partial t} + \mu\epsilon\frac{\partial^2 \mathbf E}{\partial t^2}$

$\frac{\partial^2 \mathbf E}{\partial t^2} \ll \frac{\partial \mathbf E}{\partial t}$

$\begin{align*}&\Rightarrow\widetilde k^2 = i\mu\sigma\omega\\ &\Rightarrow \widetilde k = \frac{1+i}{\sqrt{2}}\sqrt{\mu\sigma\omega}\\ &\Rightarrow \widetilde k = \kappa = \sqrt{\frac{\mu\sigma\omega}{2}}\\ \end{align*}$

Hence, we could have

$\delta = \frac{1}{k} = \sqrt{\frac{2}{\mu\sigma\omega}}$

$\delta$ is

$\left\{\begin{matrix} \text{few nm - visible light} \\ \text{few} \mu m\text{ - microwave} \\ \text{few mm - radio waves} \end{matrix}\right.$

Lets go to poor conductors

$\sigma \ll \epsilon\omega$ , hence

$k \approx \sqrt{\frac{\mu\epsilon}{2}}\omega\sqrt{2} = \sqrt{\mu\epsilon}\omega$ $\kappa = \sqrt{\frac{\mu\epsilon}{2}}\omega\left(1+\frac{1}{2}\left(\frac{\sigma}{\epsilon\omega}\right)^2+\cdots-1\right)^{\frac{1}{2}}$

$\frac{\sigma}{2}\sqrt{\frac{\mu}{\epsilon}}$ $\omega$

$\sigma = 0$ $\kappa = 0$ $k = \frac{\omega}{v}$ as expected.

Lets return to previous equation

Lets consider the curl equation in the conductor

$\begin{align*}\frac{\partial\mathbf E_x}{\partial z} &= \frac{\partial B_y}{\partial t}\\ i(k+k)\mathbf E_0 &= i\omega\mathbf B_0\\ z = \frac{\mu E_0}{B_0} = \frac{\mu\omega}{k+i\kappa}\\ \widetilde k = \frac{\mu\omega}{\sqrt{k^2+\kappa^2}}e^{i\phi}\end{align*}$

$\phi$ would be

$\phi = \tan^{-1}\left(\frac{\sqrt{1+(Q/\epsilon\omega)^2}-1}{\sqrt{1+(Q/\epsilon\omega)^2}+1}\right)^{\frac{1}{2}}$

$\sigma \gg \epsilon\omega$ $\phi \rightarrow \tan^{-1} 1 = \frac{\pi}{4}$

So this means that B lags behind E in a metal

4.4 Poynting vectors

Work done on charge

$\delta q = \rho=\delta \tau$

$\begin{align*}\mathbf\delta F &= \delta q(\mathbf{E +v\times B})\\ \mathbf{\delta F\cdot d l} &= \delta q(\mathbf{E+v\times B})\cdot \mathbf v\delta t\\ &= \mathbf{E\cdot J_f}\delta\tau\delta t\end{align*}$

$\mathbf J_f$ $\rho\mathbf v$

$\begin{align*}\mathbf F &= \frac{dw}{dt} \\ &= \mathbf{E\cdot J_f}d\tau\\ &= \frac{d}{d t}\int \underbrace{u_{mech}}_{\text{Energy density}}d\tau\end{align*}$

$\mathbf J_f = \mathbf{\nabla\times H} - \frac{\partial \mathbf D}{\partial t}$ We have

$\mathbf{E\cdot J_f} = \mathbf{E\cdot\nabla\times H} - \mathbf{E\cdot\frac{\partial D}{\partial t}}$

By dotting everything, and then

$\mathbf{\nabla\cdot(E\times H)} = \mathbf{H\cdot\nabla\times E} - \mathbf{E\cdot\nabla\times H}$ $\mathbf{E\cdot J_f = H\cdot\underbrace{\nabla\times E}_{-\frac{\partial \mathbf B}{\partial t}} - E\cdot}\frac{\partial\mathbf D}{\partial t} - \mathbf{\nabla\cdot(E\times H)}$

$\mathbf{H\cdot\underbrace{\nabla\times E}_{-\frac{\partial \mathbf B}{\partial t}} - E\cdot}\frac{\partial\mathbf D}{\partial t}$ $\frac{\partial}{\partial t}u_{EM}$ $u_{EM} = \frac{1}{2}(\mathbf{B\cdot H+E\cdot D})$ $\mathbf{\nabla\cdot(E\times H)}$ $\mathbf{S = E\times H}$ " or Poynting vector.

How is this working?

$\begin{align*} \mathbf E\cdot\frac{\partial \mathbf D}{\partial t} &= \frac{1}{2}\cdot\frac{\partial (\mathbf{E\cdot D})}{\partial t}\\ \mathbf H\cdot\frac{\partial \mathbf B}{\partial t} &= \frac{1}{2}\cdot\frac{\partial (\mathbf{B\cdot H})}{\partial t}\\ \end{align*}$ $\left.\begin{align*} \frac{1}{2}\epsilon_0E^2 &= \frac{1}{2}DE \\ \frac{B^2}{2\mu_0} &= \frac{1}{2}BH\end{align*}\right\}\text{In free space}$

$\Rightarrow \frac{d}{dt}(u_{\text{mech}}+u_{\text{EM}}) + \mathbf{\nabla\cdot S} = 0$

$\mathbf S$ is the poyting vector, or equiviantly,

$\frac{d}{dt}\int(u_{\text{mech}}+u_{\text{EM}})d\tau + \oint \mathbf S\cdot d\mathbf a = 0$

$\mathbf S$ is the energy flux density, or the rate of flow of energy per unit area in the direction of S.

Example: a capacitor

The stored energy increase at rate

$\dot{U} = \frac{Q}{C}\frac{dQ}{dt} \quad U = \frac{Q^2}{2C}$

also:

Hence, we have

$\dot{U} = S\cdot 2\pi r d = \frac{Q}{c}\frac{dQ}{dt}$

$S = EH = \frac{Q}{dc}\frac{\frac{dQ}{dt}}{2\pi r}$

There is another example

$\begin{align*}H \cdot S\pi a &= I\\ E &= \frac{V}{l}\\ \Rightarrow S &= \frac{V}{l}\frac{I}{2\pi a}\\ \int\mathbf{S\cdot da} &= -IV = -I^2R\end{align*}$

4.5 Radiation pressure

EM waves are made up of photons, and hecne they have momentum

$E = pc$

$\Rightarrow$ Transport of energy is appoinated by transport of momentum

$P_{rad} = \frac{<S>}{c}$

$P_{rad}$ is the radiation pressure

Example For a plane EM wave in free space, we have

$\begin{align*}U &= \frac{1}{2}\epsilon_0E^2+\frac{1}{2}\frac{B^2}{\mu_0}\\ \text{but } E &= cB\\ \Rightarrow U &= \epsilon_0E^2\\ \mathbf E &= \frac{1}{2}E_0\cos(kz-\omega t)\mathbf{\hat{x}}\\ &= E_0\cos^2(kz-\omega t)\\ \langle E^2\rangle &= \frac{1}{2}E_0^2\\ \Rightarrow \langle u\rangle&=\frac{1}{2}\epsilon_0E_0^2\\ \langle S\rangle &= \frac{1}{2}\epsilon_0E_0^2c = I\end{align*}$ $\Rightarrow P_{rad} = \left\{ \begin{matrix}\frac{1}{2}\epsilon_0E_0^2&& \text{perfect absorber}\\ \epsilon_0E_0^2 &&\text{perfect reflector}\end{matrix} \right.$

$1kWm^{-2}$ $\Rightarrow P_{rad} = 10^{-5}Pa$ $P_{atm} = 10^5Pa$

Example Consider a star which is growing by accretion

i.e. matter is falling onto it uniformly in all directions

$L_{sun} = 4\times 10^{26}$ w)

$\frac{L}{4\pi R^2}[Wm^{-2}]$

$\frac{L}{4\pi R^2C}$

Outward:

$k\frac{L}{4\pi R^2 C}$ where k is the opacity, which is area/unit mass, which is a constant

$\frac{GM}{R^2}$

Since they balance, we have

$\Rightarrow L = \frac{4\pi CGM}{k}$

which is called Eddington Limit (Upper limit of luminosity of stars that accrete (isotopically))

4.6 EM waves - reflection and refraction

$E_ie^{i(k_1z - \omega t)}+ E_re^{i(-k_1z-\omega t)}$

$E_te^{i(k_2z-\omega t)}$

Using electromagnetic boundary conditions, we could get

$E^\parallel$ $E^\parallel_i + E^\parallel_r = E^\parallel_t$

$H^\parallel$ $\frac{E_i}{Z_1} - \frac{E_r}{Z_1} = \frac{E_t}{Z_2}$

Putting two equations together

$\frac{E_r}{E_i} = \frac{Z_2-Z_1}{Z_2+Z_1}\quad \frac{E_t}{E_i} = \frac{2Z_2}{Z_2+Z_1}$

$Z = \sqrt{\frac{\mu}{\epsilon}}$

$= S = |\mathbf{E\times H}| = \frac{E^2}{Z}$

$S_\text{incident} = S_\text{reflected} + S_\text{transmitted}$

Where they equal to

$\frac{E_i^2}{Z_1}+\frac{E_r^2}{Z_1}=\frac{E_t^2}{Z_2}$

separately

$\mathbf E_re^{i(\mathbf{k_r\cdot r}-\omega t)}$

$\mathbf{E_i}e^{i(\mathbf{k_i\cdot r}-\omega t)}$

$\mathbf E_te^{i(\mathbf{k_t\cdot r}-\omega t)}$

$\mathbf k_i$ in x-z plane

$E_\parallel$ $\Rightarrow$ $\omega$ must be the same

$\Rightarrow \mathbf{k_i\cdot r = k_r\cdot r = k_t\cdot r}$ for all x, y at z=0

$\mathbf r = (0,y,0)$

$\Rightarrow \mathbf{k_i, k_r}$ $\mathbf k_t$ all lie in the xz plane (the plane of incidence)

$\mathbf r = (x,0,0)$ $\mathbf{k_i\cdot r} = k\sin \theta_x$

$\begin{align*}|\mathbf{k_i|=|k_r|}&=k_1\\ |\mathbf{k_t}| &= k_2\\ \Rightarrow \underbrace{k_1\sin\theta_i = k_1\sin\theta_r}_{\theta_i = \theta_r\text{, law of reflection}} &= k_2\sin\theta_t\end{align*}$

$\frac{\omega}{k}=\frac{c}{n}$ , And the last two would lead to

$\frac{\sin\theta_t}{\sin\theta_i} = \frac{k_1}{k_2} = \frac{n_2}{n_1}\quad\text{law of refraction, or snell's law}$

$n = \sqrt{\epsilon_r\mu_r}$

Fresnel equations

Worring about polarization directions

We work in those steps

$\mathbf E$ in the plane of incidence

"parallel-like" = parallel

$\mathbf{E,H}$ $\mathbf k$ form a right-handed system

	incident	reflected	transmitted
$E_x$	$E_i\cos\theta_i$	$E_r\cos\theta_r$	$E_t\cos\theta_t$
$E_z$	$-E_i\sin\theta_i$	$E_r\cos\theta_r$	$-E_t\sin\theta_t$
$H_y$	$\frac{E_i}{Z_1}$	$-\frac{E_r}{Z_1}$	$\frac{E_t}{Z_2}$

$E_\parallel$ $\Rightarrow E_x$ $\Rightarrow E_i\cos\theta_i + E_r\cos\theta_i = E_t\cos\theta_t$

$\begin{align*}\frac{E_r}{E_i} &= \frac{Z_2\cos\theta_i-Z_1\cos\theta_i}{Z_2\cos\theta_1 + Z_1\cos\theta_i}\\ \frac{E_t}{E_i} &= \frac{2Z_2\cos\theta_i}{Z_2\cos\theta_i + Z_1\cos\theta_i}\end{align*}$

Now look for Fresnel equations for p-polarizations

$\mathbf E$ perpendicular to the plane of incidence

"s-like" s = senkrecht = perpendicular

	incident	reflected	transmitted
$E_y$	$E_i$	$E_r$	$E_t$
$H_x$	$-\frac{E_i}{Z_1}\cos\theta_i$	$\frac{E_r}{Z_1}\cos\theta_r$	$-\frac{E_t}{Z_2}\cos\theta_t$
$H_z$	$\frac{E_i}{Z_1}\sin\theta_i$	$\frac{E_r}{Z_1}\sin\theta_r$	$\frac{E_t}{Z_2}\sin\theta_t$

$E_\parallel$ $\Rightarrow E_y\text{continuous} \quad E_i+E_r = E_t$ $H_\parallel$ $\Rightarrow H_x \text{continuous} \quad -\frac{E_i}{Z_1}\cos\theta_i + \frac{E_r}{Z_1}\cos\theta_r = -\frac{E_t}{Z_2}\cos\theta_t$

Remember that

$Z = \sqrt{\frac{\mu_r\mu_0}{\epsilon_r\epsilon_0}} = \frac{Z_0}{n}\quad n = \sqrt{\epsilon_r}\text{while } \mu_r = 1$

$\mu_r = 1$ $Z = \sqrt{\frac{\mu_r\mu_0}{\epsilon_r\epsilon_0}} = \frac{Z_0}{n}\quad n = \sqrt{\epsilon_r}$

$Z_i$ $\frac{1}{n_i}$ in expressions involving ratios of Z's.

e.g. Fresnel equations for p-polarization

$r = \frac{E_r}{E_i} = \frac{\frac{1}{n_2}\cos\theta_t - \frac{1}{n_1}\cos\theta_i}{\frac{1}{n_2}\cos\theta_t+\frac{1}{n_1}\cos\theta_i}$ $= \frac{n_1\cos\theta_t-n_2\cos_theta_i}{n_1\cos\theta_t+n_2\cos\theta_i}$

$n_1\sin\theta_i = n_2\sin\theta_t$ $\begin{align*} \Rightarrow r = \frac{\sin 2\theta_t - \sin 2\theta_i}{\sin 2\theta_t + \sin 2\theta_i} \\ t = \frac{4\sin_theta_t\cos\theta_i}{\sin 2\theta_t + \sin 2\theta_i} \end{align*}$

For s-polarization, we have

$\begin{align*} r = \frac{\sin(\theta_t-\theta_i)}{\sin(\theta_t+\theta_i)}\\ t = \frac{2\sin\theta_t\cos\theta_i}{\sin(\theta_t+\theta_i)} \end{align*}$

We also have

$\begin{align*}n_1\sin\theta_i &= n_2\sin\theta_t\\ \sin\theta_t &= \frac{n_1}{n_2}\sin\theta_i\\ \cos\theta_t &= \sqrt{1-\sin^2\theta_t} = \sqrt{1-\frac{n_1^2}{n_2^2}\sin^2\theta_i}\\ \end{align*}$

Fresnel equations:

$n_1\cos\theta_i$ on top and bottom

$\begin{align*}\alpha &=\frac{\cos\theta_t}{\cos\theta_i} = \frac{1}{\cos\theta_i}\sqrt{1-(\frac{n_1}{n_2}\sin\theta_i)^2}\\ \beta&=\frac{n_2}{n_1} \end{align*}$

	$E^\parallel$ (p)	$E^\perp$ (s)
$r$	$\frac{\alpha-\beta}{\alpha+\beta}$	$\frac{1-\alpha\beta}{1+\alpha\beta}$
$t$	$\frac{2\alpha}{\alpha+\beta}$	$\frac{2}{1+\alpha\beta}$

Remember, EM waves have an energy flux given by

$S = |\mathbf{E\times H}| = \frac{E^2}{Z}$

Intensity coefficients

$R = \frac{I_r}{I_i} = |r|^2 = \left\{\begin{matrix} (\frac{\alpha-\beta}{\alpha+\beta})^2 \quad (p)\\ (\frac{1-\alpha\beta}{1+\alpha\beta})^2 \quad (s) \end{matrix}\right.$

$T = \frac{I_t}{I_i} = |t|^2\frac{n_2}{n_1}\frac{\cos\theta_t}{cos\theta_i}$

$\frac{n_2}{n_1}$ $\frac{\cos\theta_t}{\cos\theta_i}$ is due to the wavesfronts at different angles.

$T = |t|^2\alpha\beta = \left\{\begin{matrix} \alpha\beta(\frac{2}{\alpha+\beta})^2 \quad (p)\\ \alpha\beta(\frac{2}{1+\alpha\beta})^2 \quad (s) \end{matrix}\right.$

Lets check in certain cases

$\theta_i = 0 \Rightarrow \alpha = 1$

$\begin{align*}r_s = r_p = \frac{1-\beta}{1+\beta} = \frac{n_1-n_2}{n_1+n_2}\\ t_s = t_p = \frac{2}{1+\beta} = \frac{2n_1}{n_1+n_2}\\ R_s = R_p = (\frac{n_1-n_2}{n_1+n_2})^2\\ T_s = T_p - \frac{n_2}{n_1}\frac{4n_1^2}{(n_1+n_2)^2} = \frac{4n_1n_2}{(n_1+n_2)^2}\end{align*}$

Example: air/glass interface

$\begin{align*}n_1 = 1\quad n_2 &= 1.5\\ r_s = r_p &= -0.2\\ T_s = T_p &= 0.8\\ R_s = R_p &= 0.04\\ T_s = T_p &= 0.96\end{align*}$ $\Rightarrow$ 4% of light is reflected, 96% is transmitted

$n_2$ $R_s = R_p = 0.074$ , which is sa problem]

Lets take another go at differnet angle

$\theta_i = 90^\circ$ $\beta >1$

$\sin\theta_i = 1\quad \cos\theta_i = 0$ $\Rightarrow \alpha \rightarrow \infty$ $\begin{align*}r_p &= 1\quad r_s = -1\\ t_p &= 0 \quad t_s = 0\end{align*}$

$\beta<1$ $\theta_i>\theta_c$ $\theta_c$ $\beta = \frac{n_2}{n_1}$

$\theta_c = \sin^{-1}\beta$ $\alpha = \frac{1}{\cos\theta_c}\sqrt{1-(\frac{\sin\theta_c}{\beta})^2} = 0$

$r_p = -1\quad r_s = 1$ $t_p = \frac{2}{\beta}\quad t_s = 2$

$\sin\theta_c = \beta$

$r_p$ "Brewster's angle" $\theta_B$ $r_p = \frac{\alpha-\beta}{\alpha+\beta} = 0$ $\alpha=\beta = \frac{n_2}{n_1}$ $\begin{align*} \frac{1}{cos\theta_B}\sqrt{1-(\frac{\sin\theta_B}{\beta})^2} &= \beta\\ \text{square both sides:}\\ \sec^2\theta_B - \frac{1}{\beta^2}\tan^2\theta_B &= \beta^2\\ (1-\frac{1}{\beta^2})\tan^2\theta_B &= \beta^2-1\\ \frac{(\beta^2-1)}{\beta^2}\tan^2\theta_B &= (\beta^2-1)\\ \Rightarrow \tan\theta_B &= \beta\end{align*}$

Method of producing polarized light

$\mathbf E$ perpendicular to the plane of incidence

Polarizing sunglasses with transmission axis vertical reduce glare because reflected light is mainly horizontally polarized.

Total internal reflection

$\begin{align*} n_1\sin\theta_i = n_2\sin\theta_t\\ \sin\theta_t = \frac{\sin\theta_i}{n_2/n_1}\\ \end{align*}$

$n_2<n_1$ $\sin\theta_t>1$

For transmitted wave, its like

$e^{i(\mathbf{k\cdot r}-\omega t)}$ $e^{i(k_2[x\sin\theta_t+z\cos\theta_t]-\omega t)}$ $z\cos\theta t = 1-\sin^2 \theta_t$ $\underbrace{e^{i(k_2x\sin\theta_t-\omega_t)}}_{wave}\underbrace{e^{-(\sin^2\theta_t-1)^{\frac{1}{2}}k_2z}}_{e^{-z/\delta}}$ $\delta = \frac{\lambda_2}{2\pi}\frac{1}{(\sin^2\theta_t-1)^{\frac{1}{2}}}$ This is called an evanescent wave

Plane travelling wave $\begin{align*}E_x &= E_0e^{i(kz-\omega t)}\\ \mathbf{\nabla\times E} &= -\mathbf{\dot{B}} = i\omega B\\ \mathbf B &= \frac{1}{i\omega}\left|\begin{matrix}\mathbf{\hat{i}}&&\mathbf{\hat{j}}&&\mathbf{\hat{k}}\\ \partial x&&\partial y&&\partial z\\ E_x &&0&&0\end{matrix}\right|\\ &=\frac{1}{i\omega}\left(\begin{matrix}0\\\partial_z E_x\\-\partial_yE_x\end{matrix}\right)\\ &= \frac{1}{i\omega}\left(\begin{matrix}0\\ikE_x\\0\end{matrix}\right)\\ &= \frac{E_x}{\omega/k}\mathbf{\hat{y}}\\ &= \frac{E_x}{c}\mathbf{\hat{y}}\end{align*}$

$\theta_i>\theta_c$ , we have

$\begin{align*} \sin\theta_t = \frac{\sin\theta_i}{n_2/n_1}>1\\ \cos\theta_t = i\sqrt{\sin^2\theta_t-1}\\ r_s = \frac{n_1\cos\theta_i-n_2\cos\theta_t}{n_1\cos\theta_i+n_2\cos\theta_t}\\ = \frac{A - iB}{A+iB}\\ \Rightarrow R_s = |r_s|^2 = \frac{A^2+B^2}{A^2+B^2} = 1\\ t_s = \frac{2n_1\cos\theta_i}{n_1\cos\theta_i+n_2\cos\theta_t} = \frac{2A}{A+iB}\end{align*}$

$\Rightarrow \mathbf E_t = \mathbf{\hat{y}}t_sE_ie^{i(k_2[x\sin\theta_t-\omega t])}e^{-z/\delta}$

$\delta = \frac{\lambda_2}{2\pi}\frac{1}{(\sin^2\theta_t-1)^{\frac{1}{2}}}$ $\begin{align*}\mathbf{\nabla\times E} &= -\dot{\mathbf{B}} = i\omega\mathbf B\\ \mathbf B &= \frac{1}{i\omega}\left|\begin{matrix}\mathbf{\hat{i}}&&\mathbf{\hat{j}}&&\mathbf{\hat{k}}\\\partial_x&&\partial_y&&\partial_z\\0&&E_t&&0\end{matrix}\right|\\ &=\frac{1}{i\omega}\left(\begin{matrix}-\partial_zE_t\\ 0 \\ \partial_xE_t\end{matrix}\right)\\ &=\left(\begin{matrix}\frac{iE_t}{\omega\delta}\\0\\k_2\sin\theta_tE_t/\omega\end{matrix}\right)\end{align*}$

$\Rightarrow B_z$ $E_y$ $\Rightarrow$ $B_x$ $E_y$ $\Rightarrow$ $<\mathbf S> = <\mathbf{E\times H^*}>$ $\mathbf H^*$ is the complex conjugate of H

Reminder on conductors $\begin{align*}\mathbf{\nabla\times(\nabla\times E) = \nabla(\nabla\cdot E)-\nabla^2E} - \frac{\partial}{\partial t}\mathbf{\nabla\times B}\\ \mu_0\mu_r\sigma\mathbf E + \mu_0\mu_r\sigma_0\sigma_r\frac{\partial \mathbf E}{\partial t} = 0\quad\text{if good conductor}\\ \nabla^2\mathbf E = \mu_0\sigma\frac{\partial \mathbf E}{\partial t}\\ -\tilde k^{2} = i\omega\mu_0\sigma\\ \Rightarrow \tilde k = k+i\kappa\quad k = \kappa = \sqrt{\frac{\mu_0\sigma\omega}{2}}\\ \mathbf{\nabla\times E = -\mathbf{\dot{B}}}\\ -\frac{\partial B_y}{\partial t} = (\mathbf{\nabla\times E})_y =\frac{\partial E_x}{\partial z}\\ i\omega B_0= = i(k+i\kappa)E_0\\ Z = \frac{E_0}{H_0} = \frac{E_0}{B_0/\mu_0} = \frac{\mu_0\omega}{k+i\kappa}\\ = \sqrt{\frac{2\mu_0\omega}{\sigma}}\frac{1}{1+i}\\ \Rightarrow Z = \sqrt{\frac{\mu_0\omega}{2\sigma}}(1-i)\end{align*}$

Reflection from a meatal surface

air	metal
$Z_1 = \sqrt{\frac{\mu_0}{\epsilon_0}}$	$Z_2 = \sqrt{\frac{\mu_0\omega}{2\sigma}}(1-i)$

$|z_2|\ll |Z_1|$ $\sigma \gg \epsilon_0\epsilon_r\omega$

$\alpha = \frac{\sqrt{\frac{\mu_0\omega}{2\sigma}}}{\sqrt{\frac{\mu_0}{\epsilon_0}}} = \sqrt{\frac{\omega\epsilon_0}{2\sigma}}\ll1$

$\begin{align*} r &= \frac{E_r}{E_i}\quad\text{normal incidence}\\ &=\frac{Z_2-Z_1}{Z_2+Z_1} \\ &= \frac{Z_2/Z_1-1}{Z_2/Z-1+1}\\ &= \frac{\alpha(1-i)-1}{\alpha(1-i)+1}\\ &\approx \frac{1-2\alpha + \cdot\cdot\cdot}{1+2\alpha+\cdot\cdot\cdot}\\ &= 1-4\alpha+O(\alpha^2)\\ &= 1-\frac{2\omega\delta}{c}+O(\alpha^2)\\ &= 1-\frac{4\pi\delta}{\lambda}+O(\alpha^2)\end{align*}$

$\delta = \sqrt{\frac{2}{\omega\mu_0\sigma}}$ $\Rightarrow$ $\Rightarrow$ metals are shiny!

4.7 Plasmas

Plasmas are neutral gas of charged particles, such as ions and free electrons (like metals)

Examples: where you can find plasma

metal
ionosphere
stars
fusion reactor
interstellar gas /intergalactic medium
supernova remnants
radio galaxies / quasars
lightening
aurorae
fire
plasma displays

$10^2 - 10^{35}kg/m^3$ $10^0-10^{13}k$

we would only focus on cold plasma

Lets consider a slab of plasma

having number density n

$\xi$

$\rightarrow$ $E = \frac{\sigma}{\epsilon_0}$ $\sigma = ne\xi$

$\begin{align*}m\ddot{\xi}& = -eE\quad\text{where} E = \frac{ne\xi}{\epsilon_0}\\ \Rightarrow \ddot{\xi}+\omega_p^2\xi&=0\quad\text{which is SHM}\\ \text{where } \omega_p^2&=\frac{ne^2}{\epsilon_0m}\end{align*}$

This is the SHM at the $\omega_p$

$\mathbf Ee^{-i\omega t}$ $v\ll c$ )

$\mathbf\xi = \mathbf\xi_0e^{-i\omega t}$

$-m\omega^2\mathbf \xi_0 = -e\mathbf E$ $\begin{align*}\mathbf P &= -ne\mathbf\xi_0=-\frac{ne^2}{m\omega^2}\mathbf E\\ &= (\epsilon_r-1)\epsilon_0\mathbf E\\ \Rightarrow \underbrace{\epsilon_r}_{\tilde n^2} &= 1-\frac{\omega_p^2}{\omega^2}\end{align*}$

$\tilde n^2$ is the refractive index, and that can be imaginary

$\omega$ $\epsilon_r$

$\omega>\omega_p$ $\tilde n$ is real, hence EM waves can propagate

$\omega<\omega_p$ $\tilde n$ is imaginary, hence EM waves cant propagate

For example, metals are shiny, but only at optical frequencies. They will transport if it is going to much shorter wavelengths.

e.g. ionosphere

AM radios would be refracted + reflected (~ 1 MHz)

FM radio and TV radios would escape (~100MHz)

Lets look at the dispersion relation again

$\frac{\omega}{k} = \frac{c}{n}$

we could get

$\begin{align*}\tilde n&= 1-\frac{\omega_p^2}{\omega^2} = \frac{c^2k^2}{\omega^2}\\ \Rightarrow \omega&=c^2k^2+\omega_p^2\end{align*}$

The dispersion relationship would be like

Hence, we could conclude that

waves are dispersive in plasma
$\omega<\omega_p$
$\omega\approx \omega_p$ $v_g = \frac{d\omega}{dk}\rightarrow0$ $\omega\rightarrow \omega_p$

$2\omega d\omega = 2c^2kdk$

$\Rightarrow \frac{d\omega}{dk} = \frac{c^2k}{\omega} = c\sqrt{1-\frac{\omega_p^2}{\omega^2}}$

We could have end behaviors:

$\begin{align*}v_g\rightarrow0\quad\text{as}\quad\omega\rightarrow\omega_p\\ v_g\rightarrow c \quad\text{as}\quad\omega\rightarrow \infty\end{align*}$

Let's then take

$\mathbf E = \mathbf E_0e^{i(\mathbf{k\cdot r}-\omega t)}$

$\mathbf k$ $(0,0,k)$ , we have

$\mathbf P = \frac{-ne^2}{m\epsilon_0\omega^2}(\epsilon_0\mathbf E) = -\left(\frac{\omega_p}{\omega}\right)^2\epsilon_0\mathbf E$

We could hence get Maxwell's equation

$\begin{align*}\mathbf{\nabla\times H} &= \frac{\partial\mathbf{\dot{D}}}{\partial t}\quad \mathbf{\nabla\times E} =-\frac{\partial \mathbf B}{\partial t} = -\mu_0\frac{\partial \mathbf H}{\partial t}\\ \mathbf{\nabla\times(\nabla\times E)} &= -\mu_0\frac{\partial}{\partial t}\mathbf{\nabla\times H}\\ &= 0\mu_0\frac{\partial^2}{\partial t^2}\mathbf D\\ \textcolor{brown}{\text{LHS} = \mathbf{\nabla(\nabla\cdot E) - \nabla^2 E}}&= -\mu_0\frac{\partial ^2}{\partial t^2}(\epsilon_0\mathbf{E+P})\end{align*}$

For the wave, we have

$\mathbf{\nabla\rightarrow}i\mathbf k$ $\frac{\partial }{\partial t}\rightarrow -\omega^2\mathbf E$ $\Rightarrow k^2\mathbf E -\underbrace{\mathbf{k(k\cdot E)}}_{k^2E_z\mathbf{\hat{z}}} = \frac{\omega^2-\omega_p^2}{c^2}\mathbf{E}$

$\left(\begin{matrix}1\\0\\0\end{matrix}\right)$ $\left(\begin{matrix}0\\1\\0\end{matrix}\right)$ $\omega^2=\omega_p^2+c^2k^2$

$\left(\begin{matrix}0\\0\\1\end{matrix}\right)$ $\omega^2=\omega_p^2$

$\omega$ k graph

4.8 Dispersion

Refractive index changes with frequency

classical theory of dispersion:

$m\ddot{x} + m\gamma\dot{x}+ m\omega_0^2x = qE_0e^{-i\omega t}$

$x = x_0e^{-i\omega t}$ $\begin{align*}(-\omega^2-i\omega\gamma+\omega_0^2)x_0 &= \frac{qE_0}{m}\\ P &= nqx_0 = \frac{nq^2}{m}E_0\frac{1}{\omega_0^2-\omega^2-i\omega\gamma}\\ &= \epsilon_0(\tilde \epsilon_r-1)E_0\\ \tilde{\epsilon_r}&= 1+\frac{nq^2E_0}{\epsilon_0m}\frac{1}{\omega_0^2-\omega-i\omega\gamma}\\ \tilde n^2&=\tilde\epsilon_r\\ \text{For a gas, } \tilde{\epsilon} &= 1+ \text{small quantity}\\ \tilde\epsilon^{\frac{1}{2}}&=1+\frac{1}{2}\text{small quantities}+\cdots\\ Re(\tilde n)&=1+\frac{nq^2}{2\epsilon_0m}\frac{\omega_0^2-\omega^2}{(\omega_0^2-\omega^2)^2+\omega^2\gamma^2}\\ Im(\tilde n)&=\frac{nq^2}{2\epsilon_0m}\frac{\gamma\omega}{(\omega_0^2-\omega^2)^2+\omega^2\gamma^2}\end{align*}$

$\omega\rightarrow 0$ $Re(\tilde n)\rightarrow 1+\frac{nq^2}{2q_0\omega_0^2}$ $Im(\tilde n)\rightarrow 0$ $\omega\rightarrow \infty$ $Re(\tilde n)\rightarrow 1$ $\tilde n$

$Im(\tilde n)$ $Re(\tilde n)$ corresponds to the refraction of light and generally increases with frequency

$\tilde n$ $\tilde n$ $\omega_0$ is called anomalous dispersion

5. Confined EM waves

5.1 Transmission lines

An example of guided wave

Lets think in a super long circuit

$\ll\frac{1}{\omega}$ , we need to consider the wave behavior of the signal.

We could have

Remember: we are defining capacitance as C per unit length, and inductance L per unit length

$\begin{align*} Q &= (C\delta z)v\\ \frac{dQ}{dt}&= C\delta z\frac{\partial v}{\partial t}\\ &=I(z,t)-I(z+dz,t)\\ &=\delta z\left(\frac{-\partial I}{\partial z}\right)\\ \Rightarrow\frac{\partial I}{\partial z}&=-C\frac{\partial v}{\partial t}\quad\text{ equation 1}\\ \Phi&=(L\delta z)I\\ \frac{d\Phi}{dt}&=L\delta z\frac{\partial I}{\partial t}\\ &=V(z,t)-V(z+\delta z,t)\\ &=\delta z\left(-\frac{\partial V}{\partial z}\right)\\ \Rightarrow \frac{\partial V}{\partial z}&=-L\frac{\partial I}{\partial t}\quad\text{equation 2} \end{align*}$

If we plug equation 1 into equation 2, we get

$\frac{\partial^2 V}{\partial t^2}=\frac{1}{LC}\frac{\partial^2 V}{\partial z^2}$

For wavelike situations, we have

$V(z,t) = f(z-vt)+g(z+vt)$

$f$ $g$ $v$ $\frac{1}{\sqrt{LC}}$ .

Note: the wave velocity is not the speed of light, but the speed of the wave in the circuit.

$\frac{d V}{d I} = \frac{1}{C\frac{d z}{d t}} = \frac{1}{C\frac{1}{\sqrt{LC}}} = \frac{1}{\sqrt{\frac{L}{C}}}$ $Z = \pm\sqrt{\frac{L}{C}}$ $\pm$ is the direction of the wave.

Instantaneous power

$\begin{align*} &= V\underbrace{(\frac{1}{2}LI^2+\frac{1}{2}CV^2)}_{\text{energy stored/length}}\\ &= \frac{1}{2}I^2z+\frac{1}{2}\frac{V^2}{z} = I^2z = \frac{LC}{z} = IV\end{align*}$

Example

We have coaxial transmission line, which has

$C = \frac{Q}{V}$

$C$ is per unit length

We would have

$E\cdot 2\pi rl = \frac{Q}{\epsilon_r\epsilon_0}$

Hence, we have

$\begin{align*} V &= -\int_b^a\frac{Q}{2\pi\epsilon_0\epsilon_r}dr = \frac{Q}{2\pi\epsilon_0\epsilon_r}\ln\frac{b}{a}\\ \Rightarrow C &= \frac{2\pi\epsilon_r\epsilon_0}{\ln\frac{b}{a}}\\ B\cdot2\pi r&=\mu_r\mu_0I\\ L &= \frac{\Phi}{I} = \frac{\mu_r\mu_0}{2\pi}\ln\frac{b}{a}\\ \end{align*}$

$\Phi$ is the flux per unit length

$\begin{align*}\Rightarrow \sigma &= \frac{1}{\sqrt{LC}} = \underbrace{\frac{1}{\sqrt{\mu_r\mu_0\epsilon_r\epsilon_0}}}_{\text{Speed of light in dielectric}}\\ Z&=\sqrt{\frac{L}{C}} = \sqrt{\frac{\mu_r\mu_0}{\epsilon_0\epsilon_r}}\frac{\ln\frac{b}{a}}{2\pi}\end{align*}$

Then, a common type of transmission line:

Example: Strip transmission line

$a\gg d$ so we could ignore edge effect

We have

$\begin{align*} C &= \frac{\epsilon_0\epsilon_r a}{d}\\ L &= \frac{\mu_0\mu_r d}{a}\\ \sigma &= \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{\mu_0\mu_r\epsilon_0\epsilon_r}}\quad\text{as before}\\ z &= \sqrt{\frac{L}{C}} = \sqrt{\frac{\mu_0\mu_r}{\epsilon_0\epsilon_r}}\frac{d}{a}\\ \end{align*}$

Boundary between two transmission lines

For x<0, we have

$V_1 = V_ie^{i(\omega t-kx)}+V_re^{i(\omega t+kx)}$

For x>0, we have

$V_2 = V_te^{i(\omega t-k_2x)}$

At x = 0, we need to match voltages

$\Rightarrow V_i+V_r = V_t\quad\text{(1)}$

Match currents, we have

$\frac{V_i-V_r}{Z_1} = \frac{V_t}{Z_2}\quad\text{(2)}$

$r = \frac{Z_2-Z_1}{Z_2+Z_1}$ $t = \frac{2Z_2}{Z_2+Z_1}$

$\frac{V_i^2}{Z_1} = \frac{V_t^2}{Z_2}+\frac{V_r^2}{Z_1}$ as expected.

Termination of load

$\frac{V_T}{I_T} = Z_T$ is the boundary condition

$r = \frac{Z_T-Z_c}{Z_T+Z_c}\quad t = \frac{2Z_T}{Z_T+Z_c}$

We would also like to consider special cases

	$Z_T$	$r$	$t$	$V_T$	$I_T$
Short circuit	0	-1	0	0	$2I_i$
Open circuit	$\infty$	1	2	$2V_i$	0
Matched	$Z_c$	0	1	$V_i$	$I_i$

$Z_T=Z_c$ , we would get maximum power transfer

proof

$= \frac{\langle V_i^2 \rangle}{Z_c} = \frac{1}{2}\frac{V_i^2}{Z_c}$

$= \frac{\langle V_T^2 \rangle}{Z_T} = \frac{1}{2}\frac{V_T^2}{Z_T} = \frac{1}{2}\frac{4Z_TV_T^2}{(Z_T+Z_c)^2}\quad\text{(3)}$ $\begin{align*}\frac{\text{Power transmitted}}{\text{incident power}} &= \frac{\text{(3)}}{\text{(4)}}\\ &=\frac{4Z_cZ_T}{(Z_T+Z_c)^2}\end{align*}$

$\begin{align*}\frac{d}{dZ_T}\left[\frac{Z_T}{(Z_T-Z_c)^2}\right] &= 0\\ \frac{(Z_T+Z_c)^2\cdot 1-2(Z_T+Z_c)Z_T}{(Z_T+Z_c)^2} &=0\\ \Rightarrow (Z_T+Z_c)\cdot(Z_T+Z_c-2Z_T) &= 0\\ \Rightarrow Z_T &= Z_c\end{align*}$

Input impedance of short sections

Input impedance

$\begin{align*} Z_{in} &=\frac{V(-l)}{I(-l)}\\ &= \frac{V_ie^{ikl}+V_re^{-ikl}}{\frac{V_i}{Z_c}e^{ikl}-\frac{V_r}{Z_c}e^{-ikl}}\\ \text{and }V_r&=rV_i \quad\text{ where}\quad r = \frac{Z_T-Z_c}{Z_T+Z_c}\\ \Rightarrow Z_{in}&=\left(\frac{Z_T\cos(kl)+iZ_c\sin(kl)}{Z_c\cos(kl)+iZ_T\cos(kl)}\right)Z_c\\ \text{For a } &\frac{\lambda}{2}\text{ line }(kl=\pi)\\ &\Rightarrow Z_{in} = Z_T\\ \text{For a } &\frac{\lambda}{4}\text{ line }(kl=\frac{\pi}{2})\\ &\Rightarrow Z_{in} = \frac{Z_c}{Z_T}\\ \end{align*}$

$Z_T$ $\Rightarrow$ $Z_{in} = iZ_c\tan(kl)$ $Z_T$ $\infty$ $\Rightarrow$ $Z_{in} = -iZ_c\cot(kl)$

$\frac{\lambda}{4}$ $Z_{in} = \frac{Z^2_c}{Z_T}$

$\frac{Z'^2_c}{Z_T} = Z_c$ $Z_c' = \sqrt{Z_cZ_t}$ The transmission line has been perfectly terninated and there is no reflected wave

Waveguides

Example of interfering EM waves

$E\parallel y$ $\mathbf k_\pm =\mathbf k_0(\pm\sin\theta,0\cos\theta)$

And we add them up

$\begin{align*} \mathbf E &= \frac{\mathbf E_0}{2}\left(e^{i(\mathbf k_+\cdot\mathbf r-\omega t)}-e^{i(\mathbf k_-\cdot\mathbf r-\omega t)}\right)\mathbf{\hat y}\\ &=\frac{\mathbf E_0}{2}e^{i(k_0z\cos\theta-\omega t)}\left(e^{ik_0x\sin\theta}-e^{-ik_0x\sin\theta}\right)\mathbf{\hat y}\\ &=\mathbf E_0 i\mathbf{\hat y}e^{i(k_0z\cos\theta-\omega t)}\sin(k_0x\sin\theta)\\ \end{align*}$

$k_0x\sin\theta=n\pi$ $x=na$ $a=\frac{\pi}{k_0\sin\theta}$

Nodal plane would be like

$\mathbf E=0$

This demonstrates (at least in 1 dimension) that guided waves are possible

When confined

$\begin{align*} \mathbf B &= \frac{1}{c}\frac{\mathbf E_0}{2}e^{i(k_0z\cos\theta-\omega t)}\left[e^{ik_0\sin\theta}\left(\begin{matrix}-\cos\theta\\0\\\sin\theta\end{matrix}\right)+e^{-ik_0\sin\theta}\left(\begin{matrix}-\cos\theta\\0\\-\sin\theta\end{matrix}\right)\right]\\ &=\frac{\mathbf E_0}{2c}e^{i(k_0z\cos\theta-\omega t)}\left(\begin{matrix}-i\cos\theta\sin(k_0x\sin\theta)\\0\\\sin\theta\cos(k_0x\sin\theta)\end{matrix}\right)\\ \end{align*}$ $\Rightarrow$ $\mathbf E$ $\mathbf B$ is not

Rectangular waveguide

Rectangular cross section metal walls

$\mathbf E$ $\mathbf B\propto e^{-i\omega t}$ $\mathbf E=\mathbf B =0$ $\mathbf B_\perp$ $\mathbf E_\parallel$ $\Rightarrow \mathbf B_\perp$ $\mathbf E_\parallel$ are zero at the walls

$\begin{align*} \mathbf{\nabla\cdot E} &=0\quad\mathbf{\nabla\times E} = i\omega\mathbf B\\ \mathbf{\nabla\cdot B} &=0\quad\mathbf{\nabla\times B} = \frac{i\omega}{c^2}\mathbf E\\ \end{align*}$

$\mathbf{\nabla\times\nabla\times E} = i\omega\mathbf{\nabla\times B} = \frac{\omega^2}{c^2}\mathbf E$ $\mathbf{\underbrace{\nabla(\nabla\cdot E)}_{0}}-\nabla^2\mathbf E = -\nabla^2\mathbf E$ , so we can have a second order differential equation

$\nabla^2\mathbf E + \frac{\omega^2}{c^2}\mathbf E = 0$

Which is called the Helmholtz equation

$\mathbf E$ $\Rightarrow E_z=0$ $\mathbf E_\parallel$ $\left\{\begin{matrix}E_x=0\quad \text{at }\quad y=0,b\\ E_y=0\quad\text{at }\quad x = 0,a\end{matrix}\right.$ $E(\mathbf r,t)\propto e^{i(k_gz-\omega t)}$ $\Rightarrow \left[\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\left(\frac{\omega^2}{c^2}-k_g^2\right)\right]\mathbf E = 0$

$E_z$ = 0

Boundary conditions $\begin{align*}E_x&\propto\sin\frac{n\pi y}{b}\\ E_y&\propto\sin\frac{m\pi x}{a}\\ E_x &= f(x)\sin\frac{n\pi y}{b}e^{i(k_gz-\omega t)}\\ E_y &= g(y)\sin\frac{m\pi y}{a}e^{i(k_gz-\omega t)}\end{align*}$

$\mathbf{\nabla\cdot E} = 0$ $\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\underbrace{\frac{\partial E_z}{\partial z}}_{0} = 0$ , hence we could continue

$\begin{align*} \frac{\partial f}{\partial x}\sin\frac{n\pi y}{b}+\frac{\partial g}{\partial y}\sin\frac{m\pi x}{a} = 0\\ \frac{\partial f}{\partial x}=-\sin\frac{m\pi x}{a}\Rightarrow f(x) = -\frac{a}{m\pi}\cos\frac{m\pi x}{a}\\ \frac{\partial g}{\partial y}=-\sin\frac{n\pi y}{b}\Rightarrow g(y) = -\frac{b}{n\pi}\cos\frac{n\pi y}{b}\\ \end{align*}$

If m = 1, n = 0, we would call it TE10 mode

$\begin{align*} E_x&=E_z = 0\\ E_y&= -A\frac{\pi}{a}\sin\frac{\pi x}{a}e^{i(k_gz-\omega t)}\\ B_x&= A\frac{k_g}{\omega}\sin\frac{\pi x}{a}e^{i(k_gz-\omega t)}\\ B_z &= i\frac{A}{\omega}\left(\frac{\pi}{a}\right)^2\cos\frac{\pi x}{a}e^{i(k_gz-\omega t)}\\ \end{align*}$

$B_z$ $\Rightarrow \frac{\omega^2}{c^2}-k_g^2-\left(\frac{m\pi}{a}\right)^2-\left(\frac{n\pi}{b}\right)^2=0$ $\Rightarrow \frac{\omega^2}{c^2} = k_g^2+\underbrace{\left(\frac{m\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}_{k_c^2}$ $k_c$ is the cutoff wave vector

There is a minumym frequency allowed allowed by the waveguide

$\begin{align*}\frac{2\omega d\omega}{c^2} = 2k_gdk_g\\ \Rightarrow V_{\text{group}} = \frac{d\omega}{dk_g} = \frac{c^2k_g}{\omega} = \frac{ck_g}{\sqrt{k_g^2+k_c^2}}\\ v_{\text{phase}} = \frac{\omega}{k_g} = \frac{c\sqrt{k_g^2+k_c^2}}{k_g}\end{align*}$

$v_{\text{group}}v_{\text{phase}} = c^2$ $\begin{align*}k_g&=k_0\cos\theta\\ k_c&=k_0\sin\theta\\ \frac{\omega}{c}&=k_0\end{align*}$

$TE_{10}$ $\theta_0$ with respect to the walls

6. Special relativity and electromagnetism

Just a brief introduction, no worries.

$ct,\mathbf{r}$ ), who has dot product with itself as

$\begin{align*}x^\mu x_\mu&= c^2t^2-r^2\\ \tau^2 &=t^2\left(1-\frac{x^2+y^2+z^2}{c^2t^2}\right)\\ &=t^2\left(1-\frac{v^2}{c^2}\right)\\ \Rightarrow\tau&=\frac{t}{\gamma}\\ \gamma&=\left(1-\frac{v^2}{c^2}\right)^{-\frac{1}{2}}\\ t&=\gamma\tau\quad\text{time dilation}\end{align*}$

$\frac{E}{c},\mathbf{p}$ ), who has dot product with itself as

$\begin{align*}p^\mu p_\mu&= \frac{E^2}{c^2}-p^2=m^2c^2\\ \text{since} E^2&=p^2c^2+m^2c^2\quad(m=\text{rest mass})\end{align*}$

$\partial_\mu$ which is written as

$\partial_\mu=\left(\frac{1}{c}\frac{\partial}{\partial t},\mathbf\nabla\right)$

We would just introduce this here

Then, we could have current density four vector

$J_\mu=(c\rho,\mathbf{J})$

We would consider charge at rest, then

$J^\mu=(c\rho,0)$

This could be "boosted" to a moving frame with speed v

$J'^\mu= (\rho'c,\mathbf J')$

$\rho' = \gamma\rho_0$ $\mathbf J'=-\gamma\rho_0\mathbf v$

Then, we could have continuity equation

$\begin{align*}\frac{\partial\rho}{\partial t}+\mathbf\nabla\cdot\mathbf J&=0\\ \partial_\mu J^\mu &=0\end{align*}$

In lab frame S, then, we could have

$+\lambda$ which is stationary
$-\lambda$ $-v$
$+u$

remember that the wire itself has no net charge

$\lambda_{\text{tot}} = +\lambda-\lambda = 0$

In the test charge frame S', we could have

Test charge is stationary
$\Rightarrow \gamma_+ = \gamma_u=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$
$v'=\frac{v+u}{1+\frac{uv}{c^2}}\Rightarrow \gamma_- = \frac{1}{\sqrt{1-\frac{c^2(v+u)^2}{(c^2+uv)^2}}} = \frac{c^2+uv}{\sqrt{(c^2-v^2)(c^2-u^2)}}=\frac{c^2+uv}{\sqrt{(c^2-v^2)(c^2-u^2)}} = \gamma_v\partial_u(1+\cfrac{vu}{c})$

$\gamma'_{\text{tot}} = \gamma_+\lambda-\gamma_-\left(-\frac{\lambda}{\gamma_v}\right)$

$\frac{-\lambda}{\gamma_v}$

$\begin{align*} \lambda'_{\text{tot}} &= \gamma_u\lambda-\lambda\gamma_u\cdot\left(1+\frac{uv}{c^2}\right)\\ &=-\lambda\gamma_u\frac{uv}{c^2}\\ E'&=\frac{\lambda'_{\text{tot}}}{2\pi\epsilon_0 r}=\frac{-\lambda uv\gamma_u}{2\pi\epsilon_0 c^2 r}\\ F'&=qE'=\frac{-\lambda uv\gamma_u q}{2\pi\epsilon_0 c^2 r}\\ F&=\frac{F'}{\gamma_u} = \frac{-(\lambda v)uq}{2\pi\epsilon_0 c^2 r}\quad\text{note that}\quad \lambda v = I\\ &=-qu\left(\frac{\mu_0I}{2\pi r}\right) \quad\text{note that}\quad \frac{\mu_0I}{2\pi r} = B\\ &=-quB\quad\text{Which is lorentz force} \end{align*}$

Lets remember Maxwell's equations

$\begin{align*} \mathbf B &= \mathbf\nabla\times\mathbf A\\ E &= -\mathbf\nabla V-\frac{\partial\mathbf A}{\partial t}\\ \text{If}\quad V&\rightarrow V-\frac{\partial\chi}{\partial t}\\ \text{and} \quad\mathbf A&\rightarrow\mathbf A+\mathbf\nabla\chi\\ \text{Where }\chi&\text{ is a scalar field}\\ A^\mu&=(\frac{V}{c},\mathbf A)\\ A_\mu&=(\frac{V}{c},-\mathbf A)\\ A_\mu &\rightarrow A_\mu-\partial_\mu\chi\quad\text{gauge transformation}\\ \end{align*}$

We would, therefore, need a new object called the field strength tensor

$\begin{align*} F_{\mu\nu}&=\partial_\mu A_\nu-\partial_\nu A_\mu\\ \text{Where }\mu,\nu&\text{ are indices}\\ F_{\mu\nu}&\rightarrow F_{\mu\nu}-\partial_\mu\partial_\nu\chi+\partial_\nu\partial_\mu\chi= F_{\mu\nu}\\ F_{\mu\nu}&=\begin{pmatrix}0&\frac{E_x}{c}&\frac{E_y}{c}&\frac{E_z}{c}\\ -\frac{E_x}{c}&0&-B_z&B_y\\ -\frac{E_y}{c}&B_z&0&-B_x\\ -\frac{E_z}{c}&-B_y&B_x&0\end{pmatrix}\\ F^{\mu\nu}&=\begin{pmatrix}0&-\frac{E_x}{c}&-\frac{E_y}{c}&-\frac{E_z}{c}\\ \frac{E_x}{c}&0&-B_z&B_y\\ \frac{E_y}{c}&B_z&0&-B_x\\ \frac{E_z}{c}&-B_y&B_z&0\end{pmatrix}\\ \end{align*}$

$\partial_\mu F^{\mu\nu} = \mu_0 J^\nu$ gives us the four Maxwell's equations. The first half would be simple to deduce, but the second half would be a bit more complicated so we are not going to do it here.

First two:

$\begin{align*} \partial_i F^{i0}&=\mu_0J^0\\ \Rightarrow \mathbf\nabla\cdot\mathbf E&=\frac{\rho}{\epsilon_0}\\ \partial_\mu F^{\mu i}&=\mu_0J^i\\ \Rightarrow \frac{1}{c^2}\mathbf E+\mathbf{\nabla\times B}&=\mu_0\mathbf J \end{align*}$

Bonus Lecture: Signals and sampling

Not in the exam

Fourier transform is a way to represent a signal as a sum of sinusoids. It is a continuous function of frequency.

$\begin{align*}\tilde f(\omega) &= A\int_{-\infty}^{\infty} f(t) e^{-i\omega t} dt\\ f(t) &= B\int_{-\infty}^\infty\tilde f(\omega)\\ AB &= \frac{1}{2\pi}\\ &\Rightarrow \text{often choose }A=B=\frac{1}{\sqrt{2\pi}}\\ \text{You can}&\text{ also choose} A=1\text{, } B = \frac{1}{2\pi}\end{align*}$

This is just a convention, and the latter would be what we use in this course or lecture.

$V(t)$ $\tilde V(\omega)$ .

band limited $\tilde V(\omega) = 0$ $|\omega| > \omega_c$ .

And we could get some examples of band limited signals from the lecture.

Look at some examples of fourier transform.

Delta function

$\begin{align*}\delta(t) &= \begin{cases}0 & t\neq 0\\ \infty & t=0\end{cases}\\ \int_{-\infty}^\infty \delta(t) dt &= 1\\ \tilde\delta(\omega) &= \int_{-\infty}^\infty \delta(t) e^{-i\omega t} dt = 1\\ \end{align*}$

That would give a unniform frequency spectrum, which is not band limited.

If you fourier transfor a delta function which is not at the origin, you would get a complex exponential.

$\delta(t-t_0) \rightarrow e^{-i\omega t_0}$

We will now think of something called the Dirac comb or sampling function.

$\begin{align*}f(t) &= \sum_{n=-\infty}^\infty \delta(t-nT)\\ \end{align*}$

$T$ .

WE could, therefore, write its fourier transform as a sum of fourier transforms of delta functions.

$\tilde f(t) = \sum_{n=-\infty}^\infty C_me^{\frac{2\pi i m t}{T}}$

$C_m$ would be

$C_m = \frac{1}{T}\int_{-T/2}^{T/2} f(t) e^{-\frac{2\pi i m t}{T}} dt$

$f(t)$ is a sum of delta functions, and only one would be nonzero in the region of integration, we would get

$C_m = \frac{1}{T}\int_{-T/2}^{T/2}\delta(t)e^{-\frac{2\pi i m t}{T}} dt = \frac{1}{T}$

$\Rightarrow \tilde f(t) = \frac{1}{T}\sum_{n=-\infty}^\infty e^{\frac{2\pi i m t}{T}}$

Now, we woule have fourier transform of the entire Dirac comb.

$\begin{align*}\tilde f(\omega) &= \int_{-\infty}^{\infty}f(t)e^{-i\omega t} dt \\&= \sum_{n=-\infty}^\infty \int_{-\infty}^\infty \delta(t-nT)e^{-i\omega t} dt \\&= \sum_{n=-\infty}^\infty e^{-i\omega nT}\end{align*}$

If you compare the two things we have, we would see that they are the same, and we would get

$\tilde f(\omega) = \frac{2\pi}{T}\sum_{k=-\infty}^\infty \delta(\omega - \frac{2\pi k}{T})$

$\frac{2\pi}{T}$ .

$V(t)$ $\frac{1}{T}$ $\Rightarrow$ $V_s(t) = V(t)\sum_{n=-\infty}^\infty \delta(t-nT)$

Since this is a product of two functions, we could write its fourier transform as a convolution of the fourier transforms of the two functions.

$\tilde V_s(\omega) = \tilde V(\omega) * \tilde f(\omega)\\$

$V_s$ , we would get

$\tilde V_s(\omega) = \int_{-\infty}^\infty V(t)\sum_{n=-\infty}^\infty \delta(t-nT)e^{-i\omega t} dt$

$V(t)$ is a function taht can be written as

$\frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde V(\bar \omega)e^{i\bar\omega t}d\bar\omega$

Now put everything together, we get

$\frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde V(\bar \omega)e^{i\bar\omega t}d\bar\omega\sum_{n=-\infty}^\infty \int_{-\infty}^\infty \delta(t-nT)e^{-i\omega t} dt$

The integration of dirac comb is just the sum of the integrations of the delta functions, which can be written as

$\frac{2\pi}{T}\sum_{k=-\infty}^\infty \delta\left(\omega-\bar\omega-\frac{2\pi k}{T}\right)$

And hence

$\tilde V_s(\omega) = \frac{1}{T}\sum_{k=-\infty}^\infty \tilde V(\omega-\frac{2\pi k}{T})$

That is just the convolution theorem tho.

$\tilde V(\omega)$ $\frac{2\pi}{T}$ .

This idea of having multiple copies of the fourier transform of the original function is important

If you think of samping a function, lets say a sine wave, you digitize it is certain points

The question is how few points do you need to digitize a signal

$T\ll \frac{1}{\omega}$ $\omega$ is the frequency of the signal.

You could, for example, sample from peak to trough, and you would get a good approximation of the signal, which should be the max frequency of sampling

$\omega_{max} = \frac{\pi}{T}$

This is known as the Nyquist frequency.

If we go back to the particular case of this signal, which is band limitecd

They would be nicely separated, not touching each others

BUT $\omega_{max} >\frac{\pi}{T}$ , then the copies would overlap, and you would not be able to tell which is which.

This is called aliasing.

This is a problem, decreasing its quality, and we would like to avoid it.

To avoid it the Nyquist frequency should be twice as big as the maximum frequency of the signal.

Lastly, lets talk about Johnson-Nyquist noise

Everything has noise, everything is fluctuating

$k = \frac{n\pi}{L}$ $\omega = ck$ $\Delta \omega = \frac{\pi c}{L}$ .

We also have

$\langle\text{Energy of one mode}\rangle = k_BT$

This is a result of the equipartition theorem.

Hence

$\frac{\langle\text{Energy}\rangle}{L} = \frac{k_BT}{L} = k_BT\frac{\Delta \omega}{c\pi}$

$\Delta \omega$

S, the mean power on the resistor on the right

$\langle\text{Power on R}\rangle = \underbrace{\frac{1}{2}k_BT\frac{\Delta\omega}{c\pi}}_{\text{Energy/length}}\times c$

which equals to

$\frac{k_BT\Delta\omega}{2\pi}$

$\langle I^2R\rangle$ named powwer

$I=\frac{V}{2R}$ $\Rightarrow\langle V^2\rangle = \frac{4k_BTR\Delta\omega}{\pi} = 4k_BTR\Delta f$ $\Delta f$ .